# The Elements of Euclid for the Use of Schools and Colleges/Appendix

APPENDIX.
This Appendix consists of a collection of important propositions which will be found useful, both as affording geometrical exercises, and as exhibiting results which are often required in mathematical investigations. The student will have no difficulty in drawing for himself the requisite figures in the cases where they are not given.
​1. The sum of the squares on the sides of a triangle is equal to twice the square on half the base, together with twice the square on the straight line which joins the vertex to the middle point of the base.
Let ABC be a triangle; and let D be the middle point of the base AB. Draw CE perpendicular to the base
meeting it at E; then B may be either in AB or in AB produced.
First, let B coincide with D; then the proposition follows immediately from I. 47.
Next, let E not coincide with D; then of the two angles ADC and BDC, one must be obtuse and one acute.
Suppose the angle ADC obtuse. Then, by II. 12, the square on AC is equal to the squares on AD, DC, together with twice the rectangle AD, DE; and, by II. 13, the square on BC together with twice the rectangle BD, DE is equal to the squares on BD, BC. Therefore, by Axiom 2, the squares on AC, BC, together with twice the rectangle BD, BE are equal to the squares on AD, DB, and twice the square on DC, together with twice the rectangle AD, DE. But AD is equal to DB. Therefore the squares on AC,BC are equal to twice the squares on AD, DC. ​2. If two chords intersect within a circle, the angle which they include is measured by half the sum of the in- tercepted arcs.
Let the chords AB and CD of a circle intersect at E; join AD.
The angle AEC is equal to the angles ADE, and DAE, by I. 32; that is, to the angles standing on the arcs AC and BD. Thus the angle AEC is equal to an angle at the circumference of the circle standing on the sum of the arcs AC and BD; and is therefore equal to an angle at the centre of the circle standing on half the sum of these arcs.
Similarly the angle CEB is measured by half the sum of the arcs CB and AD.
3. If two chords produced intersect without a circle, the angle which they include is measured by half the difference of the intercepted arcs.
Let the chords AB and CD of a circle, produced, intersect at E; join AD.
The angle ADC is equal to the angles EAD and AED, by I. 32. Thus the angle AEC is equal to the difference of the angles ADC and BAD; that is, to an angle at the circumference of the circle standing on an arc which is the
difference of AC and BD; and is therefore equal to an angle at the centre of the circle standing on half the difference of these arcs. ​4. To draw a straight line which shall touch two given circles.
Let A be the centre of the greater circle, and B the centre of the less circle. With centre A, and radius equal to the difference of the radii of the given circles, describe a circle; from B draw a straight line touching the circle
so described at C. Join AC and produce it to meet the circumference at D. Draw the radius BE parallel to AD and on the same side of AB; and join DE. Then DE shall touch both circles.
See I. 33, I. 29, and III. 16 Corollary.
Since two straight lines can be drawn from B to touch the described circle, two solutions can be obtained; and the two straight lines which are thus drawn to touch the two given circles can be shewn to meet AB, produced through B, at the same point. The construction is applicable when each of the given circles is without the other, and also when they intersect.
When each of the given circles is without the other we can obtain two other solutions. For, describe a circle with A as a centre and radius equal to the sum of the radii of the given circles; and continue as before, except that BE and AD will now be on opposite sides of AB. The two straight lines which are thus drawn to touch the two given circles can be shewn to intersect AB at the same point. ​5. To describe a circle which shall pass through three given points not in the same straight line.
This is solved in Euclid IV. 5.
6. To describe a circle which shall pass through two given points on the same side of a given straight line, and touch that straight line.
Let A and B be the given points; join AB and produce it to meet the given straight line at C. Make a square equal to the rectangle CA, CB (II. 14), and on the
given straight line take CE equal to a side of this square.
Describe a circle through A, B, E (5); this will be the circle required (III. 37).
Since E can be taken on either side of C, there are two solutions.
The construction fails if AB is parallel to the given straight line. In this case bisect AB at D, and draw DC at right angles to AB, meeting the given straight line at C. Then describe a circle through A, B, C.
7. To describe a circle which shall pass through a given point and touch two given straight lines.
Let A be the given point; produce the given straight lines to meet at B, and join AB. Through B draw a straight line, bisecting that angle included by the given straight lines within which A lies; and in this bisecting straight line take any point C. From C draw a perpendicular on one of the given straight lines, meeting it at D; with centre C, and radius CD, describe a circle, meeting AB, produced if necessary, at E. Join CE; and through A draw a straight line parallel to CE, meeting BC, produced if ​ necessary, at F. The circle described from the centre F, with radius FA, will touch the given straight lines.
For, draw a perpendicular from F on the straight line BD, meeting it at G. Then CE is to FA as BC is to BF, and CD is to FG as BC is to BF (VI. 4, V. 16). Therefore CE is to FA as CD is to FG (V. 11). Therefore CE is to CD as FA is to FG (V. 16). But CE is equal to CD; therefore FA is equal to FG (V. A).
If A is on the straight line BC we determine E as before; then join ED, and draw a straight line through A parallel to ED meeting BD produced if necessary at G; from G draw a straight line at right angles to BG, and the point of intersection of this straight line with BC produced if necessary, is the required centre.
As the circle described from the centre C, with the radius CD, will meet AB at two points, there are two solutions.
If A is on one of the given straight lines, draw from A a straight line at right angles to this given straight line; the point of intersection of this straight line with either of the two straight lines which bisect the angles made by the given straight lines may be taken for the centre of the required circle.
If the two given straight lines are parallel, instead of drawing a straight line BC to bisect the angle between them, we must draw it parallel to them, and equidistant from them. ​8. To describe a circle which shall touch three given straight lines, not more than two of which are parallel.
Proceed as in Euclid IV. 4. If the given straight lines form a triangle, four circles can be described, namely, one as in Euclid, and three others each touching one side of the triangle and the other two sides produced. If two of the given straight lines are parallel, two circles can be described, namely, one on each side of the third given straight line.
9. To describe a circle which shall touch a given circle, and touch a given straight line at a given point.
Let A be the given point in the given straight line, and C be the centre of the given circle. Through C draw a straight line perpendicular to the given straight line,
and meeting the circumference of the circle at B and D, of which D is the more remote from the given straight line. Join AD, meeting the circumference of the circle at E. From A draw a straight line at right angles to the given straight line, meeting CE produced at F. Then F shall be the centre of the required circle, and FA its radius.
For the angle AEF is equal to the angle CED (I. 15); and the angle EAF is equal to the angle CDE (I. 29); therefore the angle AEF is equal to the angle EAF; therefore AF is equal to EF {I. 6). ​In a similar manner another solution may be obtained by joining AB. If the given straight line falls without the given circle, the circle obtained by the first solution touches the given circle externally, and the circle obtained by. the second solution touches the given circle internally. If the given straight line cuts the given circle, both the circles obtained touch the given circle externally.
10. To describe a circle which shall pass through two given points and touch a given circle.
Let A and B be the given points. Take any point C on the circumference of the given circle, and describe a circle through A, B, C. If this described circle touches the given circle, it is the required circle. But if not, let D
be the other point of intersection of the two circles. Let AB and CD be produced to meet at E; from E draw a straight line touching the given circle at F. Then a circle described through A, B, F shall be the required circle.
See III. 35 and III. 37.
There are two solutions, because two straight lines can be drawn from E to touch the given circle.
If the straight line which bisects AB at right angles passes through the centre of the given circle, the construction fails, for AB and CD are parallel. In this case F must be determined by drawing a straight line parallel to AB so as to touch the given circle. ​11. To describe a circle which shall touch two given straight lines and a given circle.
Draw two straight lines parallel to the given straight lines, at a distance from them equal to the radius of the given circle, and on the sides of them remote from the centre of the given circle. Describe a circle touching the straight lines thus drawn, and passing through the centre of the given circle (7). A circle having the same centre as the circle thus described, and a radius equal to the excess of its radius over that of the given circle, will be the required circle.
Two solutions will be obtained, because there are two solutions of the problem in 7; the circles thus obtained touch the given circle externally.
We may obtain two circles which touch the given circle internally, by drawing the straight lines parallel to the given straight lines on the sides of them adjacent to the centre of the given circle.
12. To describe a circle which shall pass through a given point and touch a given straight line and a given circle.
We wall suppose the given point and the given straight line without the circle; other cases of the problem may be treated in a similar manner.
Let A be the given point, and B the centre of the given circle. From B draw a perpendicular to the given straight line, meeting it at C, and meeting the circumference of the given circle at D and E, so that D is between B and C. Join EA and determine a point F in EA produced if necessary, such that the rectangle EA, EF may be equal to the rectangle EC,ED; this can be done by describing a circle through A,C,D, which will meet EA at the required point (III. 36, Corollary). Describe a circle to pass through A and F and touch the given straight line (6); this shall be the required circle. ​For, let the circle thus described touch the given straight line at G; join EG meeting the given circle at H,
and join DH. Then the triangles EHD and EGG are similar; and therefore the rectangle EC, ED is equal to the rectangle EG, EH (III. 31, VI. 4, VI. 16). Thus the rectangle EA, EF is equal to the rectangle EH, EG; and therefore H is on the circumference of the described circle (III. 36, Corollary). Take K the centre of the described circle; join KG, KH, and BH. Then it may be shewn that the angles KHG and EHB are equal (I. 29, I. 5). Therefore KHB is a straight line; and therefore the described circle touches the given circle.
Two solutions will be obtained, because there are two solutions of the problem in 6; the circles thus described touch the given circle externally.
By joining DA instead of EA we can obtain two solutions in which the circles described touch the given circle internally. ​13. To describe a circle which shall touch a given straight line and two given circles.
Let A be the centre of the larger circle and B the centre of the smaller circle. Draw a straight line parallel to the given straight line, at a distance from it equal to the radius of the smaller circle, and on the side of it remote from A. Describe a circle with A as centre, and radius equal to the difference of the radii of the given circles. Describe a circle which shall pass through B, touch externally the circle just described, and also touch the straight line which has been drawn parallel to the given straight line (12). Then a circle having the same centre as the second described circle, and a radius equal to the excess of its radius over the radius of the smaller given circle, will be the required circle.
Two solutions will be obtained, because there are two solutions of the problem in 12; the circles thus described touch the given circles externally.
We may obtain in a similar manner circles which touch the given circles internally, and also circles which touch one of the given circles internally and the other externally.
14. Let A be the centre of a circle, and B the centre of a larger circle; let a straight line be drawn touching the former circle at C and the latter circle at D, and meeting AB produced through A at T. From T draw any straight line meeting the smaller circle at K and L, and the larger circle at M and N; so that the five letters T, K, L, M, N are in this order. Then the straight lines AK, KC, CL, LA shall be respectively parallel to the straight lines BM, MD, DN, NB; and the rectangle TK, TN shall he equal to the rectangle TL, TM, and equal to the rectangle TC, TD.
Join AC, BD. Then the triangles TAC and TBD are ​equiangular; and therefore TA is to TB as AC is to BD (VI. 4, V. 16), that is, as AK is to BM.
Therefore the triangles TAK and TBM are similar (YI. 7); therefore the angle TAK is equal to the angle TBM; and therefore AK is parallel to BM. Similarly AL is parallel to BN. And because AK is parallel to BM and AG parallel to BD the angle CAK is equal to the angle DBM; and therefore the angle CLK is equal to the angle DNM (III. 20); and therefore CL is parallel to DN. Similarly CK is parallel to DM.
Now TM is to TD as TD is to TN (III. 37, VI. 16); and TM is to TD as TK is to TC (VI. 4); therefore TK is to TC as TD is to TN; and therefore the rectangle TK, TN is equal to the rectangle TC, TD, Similarly the rectangle TL, TM is equal to the rectangle TC, TD.
If each of the given circles is without the other we may suppose the straight line which touches both circles to meet AB at T between A and B, and the above results will all hold, provided we interchange the letters K and L; so that the five letters are now to be in the following order, L, K, T, M, N
The point T is called a centre of similitude of the two circles. ​15. To describe a circle which shall pass through a given point and touch two given circles.
Let A be the centre of the smaller circle and B the centre of the larger circle; and let E be the given point.
Draw a straight line touching the former circle at G and the latter at D, and meeting the straight line AB, produced through A, at T. Join TE and divide it at F so that the rectangle TE, TF may be equal to the rectangle TC, TD. Then describe a circle to pass through E and F and touch either of the given circles (10); this shall be the required circle. ​For suppose that the circle is described so as to touch the smaller given circle; let G be the point of contact; we have then to shew that the described circle will also touch the larger given circle. Join TG, and produce it to meet the larger given circle at H. Then the rectangle TG,TH is equal to the rectangle TC, TD (14); therefore the rectangle TG, TH is equal to the rectangle TE, TF and therefore the described circle passes through H.
Let O be the centre of this circle, so that OGA is a straight line; we have to shew that OHB is a straight line.
Let TG intersect the smaller circle again at K; then AK is parallel to BH(14); therefore the angle AKT is equal to the angle BHG; and the angle AKG is equal to the angle AGK, which is equal to the angle OGH, which is equal to the angle OHG. Therefore the angles BHG and OHG together are equal to AKT and AKG together; that is, to two right angles. Therefore OHB is a straight line.
Two solutions will be obtained, because there are two solutions of the problem in 10. Also, if each of the given circles is without the other, two other solutions can be obtained by taking for T the point between A and B where a straight line touching the two given circles meets B. The various solutions correspond to the circumstance that the contact of circles may be external or internal
16. To describe a circle which shall touch three given circles.
Let A be the centre of that circle which is not greater than either of the other circles; let B and C be the centres of the other circles. With centre B, and radius equal to the excess of the radius of the circle with centre B over the radius of the circle with centre A, describe a circle. Also with centre C, and radius equal to the excess of the radius of the circle with centre C over the radius of the circle with centre A, describe a circle. Describe a circle to touch externally these two described circles and to pass through A (15). Then a circle having the same centre as the last described circle, and having a radius equal to ​the excess of its radius over the radius of the circle with centre A will touch externally the three given circles.
In a similar way we may describe a circle touching internally the three given circles, or touching one of them externally and the two others internally, or touching one of them internally and the two others externally.
17. In a given indefinite straight line it is required to find a point such that the sum of its distances from two given points on the same side of the straight line shall be the least possible.
Let A and B be the two given points. From A draw a perpendicular to the given straight line meeting it at C and produce AC to D so that CD may be equal to AC Join DB meeting the given straight line at E. Then shall be the required point.
For, let F be any other point in the given straight line. Then, because AG is equal to DC, and EC is common to the two triangles ACE, DCE; and that the right angle ACE is equal to the right angle DCE; therefore AE is equal to DE. Similarly, AF is equal to DF. And the sum of DF and FB is greater than BD (I. 20): therefore the sum of AF and FB is greater than BD; that is, the sum of AF and FB is greater than the sum of DE and EB; therefore the sum of AF and FB is greater than the sum of AE and EB. ​18. The perimeter of an isosceles triangle is less than that of any other triangle of equal area standing on the same base.
Let ABC be an isosceles triangle; AQC any other triangle equal in area and standing on the same base AC.
Join BQ; then BQ is parallel to AC (I. 39).
And it will follow from 17 that the sum of AQ and QC is greater than the sum of AB and BC.
19. If a polygon he not equilateral a polygon may he found of the same number of sides, and equal in area, but having a less perimeter.
For, let CD, DE be two adjacent unequal sides of the polygon. Join CE. Through D draw a straight line parallel to CE. Bisect CE at L; from L draw a straight line at right angles to CE meeting the straight line drawn through D at K. Then by removing from the given polygon the triangle CDE and applying the triangle CKE we obtain a polygon having the same number of sides as the given polygon, and equal to it in area, but having a less perimeter (18). ​20. A and B are two given points on the same side of a given straight line, and AB produced meets the given straight line at C; of all points in the given straight line on each side of C, it is required to determine that at which AB subtends the greatest angle.
Describe a circle to pass through A and B, and to touch the given straight line on that side of C which is to be considered (6). Let D be the point of contact: D shall be the required point.
For, take any other point E in the given straight line, on the same side of C as D is; draw EA, EB; then one at least of these straight lines will cut the circumference ADB.
Suppose that BE cuts the circumference at F; join AF. Then the angle AFB is equal to the angle ADB (III. 21); and the angle AFB is greater than the angle AEB (I. 16); therefore the angle ADB is greater than the angle AEB.
21. A and B are two given points within a circle; and AB is drawn and produced both ways so as to divide the whole circumference into two arcs; it is required to determine the point in each of these arcs at which AB subtends the greatest angle. ​Describe a circle to pass through A and B and to touch the circumference considered (10): the point of contact will be the required point. The demonstration is similar to that in the preceding proposition.
22. A and B are two given points without a given circle; it is required to determine the points on the circumference of the given circle at which AB subtends the greatest and least angles.
Suppose that neither AB nor AB produced cuts the given circle.
Describe two circles to pass through A and B, and to touch the given circle (10): the point of contact of the circle which touches the given circle externally will be the point where the angle is greatest, and the point of contact of the circle which touches the given circle internally will be the point where the angle is least. The demonstration is similar to that in 20.
If AB cuts the given circle, both the circles obtained by 10 touch the given circle internally; in this case the angle subtended by AB at a point of contact is less than the angle subtended at any other point of the circumference of the given circle which is on the same side of AB. Here the angle is greatest at the points where AB cuts the circle, and is there equal to two right angles.
If AB produced cuts the given circle, both the circles obtained by 10 touch the given circle externally; in this case the angle subtended by AB at a point of contact is greater than the angle subtended at any other point of the circumference of the given circle which is on the same side of AB. Here the angle is least at the points where AB produced cuts the circle, and is there zero. ​23. If there be four magnitudes such that the first is to the second as the third is to the fourth; then shall the first together with the second be to the excess of the first above the second as the third together with the fourth is to the excess of the third above the fourth.
For, the first together with the second is to the second as the third together with the fourth is to the fourth (V. 18). Therefore, alternately, the first together with the second is to the third together with the fourth as the second is to the fourth (V. 16).
Similarly, by V. 17 and V. 16, the excess of the first above the second is to the excess of the third above the fourth as the second is to the fourth. Therefore, by V. 11, the first together with the second is to the excess of the first above the second as the third together with the fourth is to the excess of the third above the fourth.
24. The straight lines drawn at right angles to the sides of a triangle from the points of bisection of the sides meet at the same point.
Let ABC be a triangle; bisect BC at D, and bisect CA at E; from D draw a straight line at right angles to BC, and from E draw a straight line at right angles to CA;
let these straight lines meet at G: we have then to shew that the straight line which bisects AB at right angles also passes through G. From the triangles BDG and ​CDG we can shew that BG is equal to CG; and from the triangles CEG and AEG we can shew that CG is equal to AG; therefore BG is equal to AG. Then if we draw a straight line from G to the middle point of AB we can show that this straight line is at right angles to AB: that is, the line which bisects AB at right angles passes through G.
25. The straight lines drawn from the angles of a triangle to the points of bisection of the opposite sides meet at the same point.
Let ABC be a triangle; bisect BC at D, bisect CA at E, and bisect AB at F; join BE and CF meeting at G; {page contains image}} join AG and GD: then AG and GD shall lie in a straight line.
The triangle BEA is equal to the triangle BEC, and the triangle GEA is equal to the triangle GEC (I. 38); therefore, by the third Axiom, the triangle BGA is equal to the triangle BGC.
Similarly, the triangle CGA is equal to the triangle CGB. Therefore the triangle BGA is equal to the triangle CGA. And the triangle BGB is equal to the triangle CGB (1. 38); therefore the triangles BGA and BGD together are equal to the triangles CGA and CGD together. Therefore the. triangles BGA and BGD together are equal to half the. triangle ABC. Therefore G must fall on the straight line AB; that is, AG and GD lie in a straight line. ​26. The straight lines which bisect the angles of a triangle meet at the same point.
Let ABC be a triangle; bisect the angles at B and C
by straight lines meeting at G; join AG: then AG shall bisect the angle at A.
From G draw GD perpendicular to BC, GE perpendicular to CA, and GF perpendicular to AB.
From the triangles BGF and BGD we can shew that GF is equal to GD, and from the triangles CGE and CGD we can shew that GE is equal to GD; therefore GF is equal to GE. Then from the triangles AFG and AEG we can shew that the angle FAG is equal to the angle EAG.
The theorem may also be demonstrated thus. Produce AG to meet BC at H. Then AB is to BH as AG is to GH, and AC is to CH as AG is to GH (VI. 3); therefore AB is to BH as AC is to CH(V.11); therefore AB is to AC as BH is to CH (V. 16); therefore the straight line AH bisects the angle at A (VI. 3).
27. Let two sides of a triangle he produced through the base; then the straight lines which bisect the two exterior angles thus formed, and the straight line which bisects the vertical angle of the triangle, meet at the same point.
This may be shewn like 26: if we adopt the second method we shall have to use VI. A. ​28. The perpendiculars drawn from the angles of a triangle on the opposite sides meet at the same point.
Let ABC be a triangle; and first suppose that it is not obtuse angled. From B draw BE perpendicular to CA;
from C draw CF perpendicular to AB; let these perpendiculars meet at G; join AG, and produce it to meet BC at D: then AD shall be perpendicular to BC.
For a circle will go round AEGF {Note on III. 22); therefore the angle FAG is equal to the angle FEG (III. 21). And a circle will go round BCEF (IV. 31, Note on III. 21); therefore the angle FEB is equal to the angle FCB. Therefore the angle BAD is equal to the angle BCF. And the angle at B is common to the two triangles BAD and BCF. Therefore the third angle BDA is equal to the third angle BFC (Note on I. 32). But the angle BFC is a right angle, by construction; therefore the angle BDA is a right angle.
In the same way the theorem may be demonstrated when the triangle is obtuse angled. Or this case may be deduced from what has been already shewn. For suppose the angle at A obtuse, and let the perpendicular from B on the opposite side meet that side produced at E, and let the perpendicular from C on the opposite side meet that side produced at F; and let BE and CF be produced to meet at G. Then in the triangle BGG the perpendiculars BF and CE meet at A; therefore by the former case the straight line GA produced will be perpendicular to BC. ​29. If from any point in the circumference of the circle described round a triangle perpendiculars be drawn to the sides of the triangle, the three points of intersection are in the same straight line.
Let ABC be a triangle, P any point on the circumference of the circumscribing circle; from P draw PD,
PE,PF perpendiculars to the sides BC, CA,AB respectively: D, B, F shall be in the same straight line.
[We will suppose that P is on the arc cut off by AB, on the opposite side from C, and that B is on CA produced through A; the demonstration will only have to be slightly modified for any other figure.]
A circle will go round PEAF (Note on III. 22); therefore the angle PFE is equal to the angle PAE (III. 21). But the angles PAE and PAC are together equal to two right angles (I. 13); and the angles PAC and PBC are together equal to two right angles (III. 22). Therefore the angle PAB is equal to the angle PBC; therefore the angle PFE is equal to the angle PBC.
Again, a circle will go round PFDB {Note on III. 21); therefore the angles PFD and PBD are together equal to two right angles (III. 22). But the angle PBD has been shewn equal to the angle PFE. Therefore the angles PFD and PFE are together equal to two right angles. Therefore EF and FD are in the same straight line. ​30. ABC is a triangle, and O is the point of intersection of the perpendiculars from A, B, C on the opposite sides of the triangle: the circle which passes through the middle points of A, OB, OC will pass through the feet of the perpendiculars and through the middle points of the sides of the triangle.
Let D, E, F be the middle points of OA, OB, OC respectively; let G be the foot of the perpendicular from A on BC, and H the middle point of BC.
Then OBC is a right-angled triangle and E is the middle point of the hypotenuse OB; therefore EG is equal to EO; therefore the angle EGO is equal to the angle EOG. Similarly, the angle FGO is equal to the angle FOG. Therefore the angle FGE is equal to the angle FOE. But the angles FOE and BAG are together equal to two right angles; therefore the angles FGE and BAC are together equal to two right angles. And the angle BAC is equal to the angle EDF, because ED, DF are parallel to BA, AG (VI. 2). Therefore the angles FGE and EDF are together equal to two right angles. Hence G is on the circumference of the circle which passes through D, E, F {Note on III. 22).
Again, FH is parallel to OB, and EH parallel to 0C therefore the angle EHF is equal to the angle EGF. Therefore H is also on the circumference of the circle. ​Similarly, the two points in each of the other sides of the triangle ABC may be shewn to be on the circumference of the circle.
The circle which is thus shewn to pass through these nine points may be called the Nine points circle: it has some curious properties, of which we will now give two.
The radius of the Nine points circle is half of the radius of the circle described round the original triangle.
For the triangle DEF has its sides respectively halves of the sides of the triangle ABC, so that the triangles are similar. Hence the radius of the circle described round DEF is half of the radius of the circle described round ABC.
If S be the centre of the circle described round the triangle ABC, the centre of the Nine points circle is the middle point ofSO.
For HS is at right angles to BC, and therefore parallel to GO. Hence the straight line which bisects HG at right angles must bisect SO. And H and G are on the circumference of the Nine points circle, so that the straight line which bisects HG at right angles must pass through the centre of the Nine points circle. Similarly, from the other sides of the triangle ABC two other straight lines can be obtained, which pass through the centre of the Nine points circle and also bisect SO. Hence the centre of the Nine points circle must coincide with the middle point of SO.
We may state that the Nine points circle of any triangle touches the inscribed circle and the escribed circles of the triangle: a demonstration of this theorem will be found in the Plane Trigonometry, Chapter xxiv. For the history of the theorem see the Nouvelles Annales de Mathematiques for 1863, page 562.
31. If two straight lines bisecting two angles of a triangle and terminated at the opposite sides be equal, the bisected angles shall be equal.
Let ABC be a triangle; let the straight line BD bisect the angle at B, and be terminated at the side AC; and let the straight line CE bisect the angle at C, and be terminated at the side AB; and let the straight line BD be equal to the straight line CE: then the angle at B shall be equal to the angle at C ​
For, let BB and GE meet at; then if the angle OBC be not equal to the angle OCB one of them must be greater than the other; let the angle OBC be the greater. Then, because CB and BD are equal to BC and CE, each to each; but the angle CBD is greater than the angle BCE; therefore CD is greater than BE (I. 24).
On the other side of the base BC make the triangle BCF equal to the triangle CBE, so that BF may be equal to CE, and CF equal to BE (I. 22); and join DF.
Then because BF is equal to BD the angle BED is equal to the angle BDF. And. the angle OCD is, by hypothesis, less than the angle OBE; and the angle COD is equal to the angle BOE; therefore the angle ODC is greater than the angle OEB (I, 32), and therefore the angle ODC is greater than the angle BFC.
Hence, by taking away the equal angles BDF and BFD, the angle FDC is greater than the angle DFC; and therefore CF is greater than CD (I. 1 9); therefore BE is greater than CD.
But it was shewn that CD is greater than BE which is absurd.
Therefore the angles OBC and OCB are not unequal, that is, they are equal; and therefore the angle ABC is equal to the angle ACB.
[For the history of this theorem see Lady's and Gentleman's Diary for 1869, page 88.] ​32. If a quadrilateral figure does not admit of having a circle described round it, the sum of the rectangles contained by the opposite sides is greater than the rectangle contained by the diagonals.
Let ABCD be a quadrilateral figure which does not admit of having a circle described round it; then the rectangle AB, DC, together with the rectangle BC, AD, shall be greater than the rectangle AC, BD.
For, make the angle ABE equal to the angle DBC, and the angle BAE equal to the angle BDC; then the triangle ABE is similar to the triangle BDC (VI. 4); therefore AB is to AE as DB is to DC; and therefore the rectangle AB, DC is equal to the rectangle AE, DB.
Join EC. Then, since the angle ABE is equal to the angle DBC, the angle CBE is equal to the angle DBA. And because the triangles ABE and DBC are similar, AB is to DB as BE is to BC; therefore the triangles ABD and EBC are similar (VI. 6); therefore CB is to CE as DB is to DA; and therefore the rectangle CB, DA is equal to the rectangle CE, DB.
Therefore the rectangle AB, DC, together with the rectangle BC, AD is equal to the rectangle AE, BD together with the rectangle CE, BD; that is, equal to the rectangle contained by BD and the sum of AE and EC. But the sum of AE and EC is greater than AC (I. 20); therefore the rectangle AB, DC, together with the rectangle BC, AD is greater than the rectangle AC, BD. ​33. If the rectangle contained by the diagonals of a quadrilateral he equal to the sum of the rectangles con-tained by the opposite sides, a circle can be descrihed round the quadrilateral.
This is the converse of VI. D; it can be demonstrated indirectly with the aid of 32.
34. It is required to find a point in a given straight line, such that the rectangle contained hy its distances from two given points in the straight line may he equal to the rectangle contained by its distances from two other given points in the straight line.
Let A, B, C, D be four given points in the same straight line: it is required to find a point in the straight
line, such that the rectangle contained by its distances from A and B may be equal to the rectangle contained by its distances from C and D.
On AD describe any triangle AED; and on CB describe a similar triangle CFB, so that CF is parallel to AE, and BF to DE; join EF, and let it meet the given straight line at O, Then shall be the required point.
For, OE is to OA as OE is to OC (VI. 4); therefore OE is to OF as OA is to OC (V. 16). Similarly OE is to OF as OD is to OB. Therefore OA is to OC as OD is to OB (V. 11). Therefore the rectangle OA, OB is equal to the rectangle OC, OD. ​The figure will vary slightly according to the situation of the four given points, but corresponding to an assigned situation there will be only one point such as is required. For suppose there could be such a point P, besides the point O which is determined by the construction given above; and that the points are in the order A,C,D, B, O, P. Join PE, and let it meet CF, produced at G; join BG. Then the rectangle PA, PB is, by hypothesis, equal to the rectangle PC, PD; and therefore PA is to PC as PD is to PB. But PA is to PC as PE is to PG ( VI. 2); therefore PD is to PB as PE is to PG (V. 11); therefore BG is parallel to DE.
But, by the construction, BF is parallel to ED; therefore BG and BF are themselves parallel (I, 30); which is absurd. Therefore P is not such a point as is required.
ON GEOMETRICAL ANALYSIS.
35. The substantives analysis and synthesis, and the corresponding adjectives analytical and synthetical, are of frequent occurrence in mathematics. In general analysis means decomposition, or the separating a whole into its parts, and synthesis means composition, or making a whole out of its parts. In Geometry however these words are used in a more special sense. In synthesis we begin with results already established, and end with some new result; thus, by the aid of theorems already demonstrated, and problems already solved, we demonstrate some new theorem, or solve some new problem. In analysis we begin with assuming the truth of some theorem or the solution of some problem, and we deduce from the assumption consequences which we can compare with results already established, and thus test the validity of our assumption.
36. The propositions in Euclid's Elements are all exhibited synthetically; the student is only employed in examining the soundness of the reasoning by which each successive addition is made to the collection of geometrical truths already obtained; and there is no hint given as to the manner in which the propositions were originally discovered. Some of the constructions and demonstrations appear rather artificial, and we are thus naturally induced to enquire whether any 'rules can be discovered by which we may be guided easily and naturally to the investigation of new propositions. ​37. Geometrical analysis has sometimes been described in language which might lead to the expectation that directions could be given which would, enable a student to proceed to the demonstration of any proposed theorem, or the solution of any proposed problem, with confidence of success; but no such directions can be given. We will state the exact extent of these directions. Suppose that a new theorem is proposed for investigation, or a new problem for trial. Assume the truth of the theorem or the solution of the problem, and deduce consequences from this assumption combined with results which have been already established. If a consequence can be deduced which contradicts some result already established, this amounts to a demonstration that our assumption is inadmissible; that is, the theorem is not true, or the problem cannot be solved. If a consequence can be deduced which coincides with some result already established, we cannot say that the assumption is inadmissible; and it may happen that by starting from the consequence which we deduced, and retracing our steps, we can succeed in giving a synthetical demonstration of the theorem, or solution of the problem. These directions however are very vague, because no certain rule can be prescribed by which we are to combine our assumption with results already established; and moreover no test exists by which we can ascertain whether a valid consequence which we have drawn from an assumption will enable us to establish the assumption itself. That a proposition may be false and yet furnish consequences which are true, can be seen from a simple example. Suppose a theorem were proposed for investigation in the following words; one angle of a triangle is to another as the side opposite to the first angle is to the side opposite to the other. If this be assumed to be true we can immediately deduce Euclid's result in I. 19; but from Euclid's result in I. 19 we cannot retrace our steps and establish the proposed theorem, and in fact the proposed theorem is false.
Thus the only definite statement in the directions respecting Geometrical analysis is, that if a consequence can be deduced from an assumed proposition which contradicts a result already established, that assumed proposition must be false. ​38. We may mention, in particular, that a consequence would contradict results already established, if we could shew that it would lead to the solution of a problem already given up as impossible. There are three famous problems which are now admitted to be beyond the power of Geometry; namely, to find a straight line equal in length to the circumference of a given circle, to trisect any given angle, and to find two mean proportionals between two given straight lines. The grounds on which the geometrical solution of these problems is admitted to be impossible cannot be explained without a knowledge of the higher parts of mathematics; the student of the Elements may however be content with the fact that innumerable attempts have been made to obtain solutions, and that these attempts have been made in vain.
The first of these problems is usually referred to as the Quadrature of the Circle. For the history of it the student should consult the article in the English Cyclopædia under that head, and also a series of papers in the Athenæum for 1863 and subsequent years, entitled a Budget of Paradoxes, by Professor De Morgan.
For approximate solutions of the problem we may refer to Davies's edition of Hutton's Course of Mathematics Vol. i. page 400, the Lady's and Gentleman's Diary for 1855, page 86, and the Philosophical Magazine for April, 1862.
The third of the three problems is often referred to as the Duplication of the Cube. See the note on VI. 13 in Lardner's Euclid, and a dissertation by C. H. Biering entitled Historia Problematis Cubi Duplicandi... Hauniæ, 1844.
We will now give some examples of Geometrical analysis.
39. From two given points it is required to draw to the same point in a given straight line, two straight lines equally inclined to the given straight line.
Let A and B be the given points, and CD the given straight line.
Suppose AE and EB to be the two straight lines equally inclined to CD. Draw BF perpendicular to CD, and produce AE and BF to meet at G. Then the angle ​BED is equal to the angle AEC by hypothesis; and the angle AEC is equal to the angle DEG (I. 15). Hence the
triangles BEF and GEF are equal in all respects (I. 26); therefore FG is equal to FB.
This result shews how we may synthetically solve the problem. Draw BF perpendicular to CD, and produce it to G, so that FG may be equal to FB; then join AG, and AG will intersect CD at the required point.
40. To divide a given straight line into two parts such that the difference of the squares on the parts may be equal to a given square.
Let AB he the given straight line, and suppose C the required point.
Then the difference of the squares on C and BG is to be equal to a given square. But the difference of the squares on AC and BC is equal to the rectangle contained by their sum and difference; therefore this rectangle must be equal to the given square. Hence we have the following synthetical solution. On AB describe a rectangle equal to the given square (I. 45); then the difference of AC and CB will be equal to the side of the rectangle adjacent to AB, and is therefore known. And the sum of AC and CB is known. Thus AC and CB are known.
It is obvious that the given square must not exceed the square on AB, in order that the problem may be possible.
There are two positions of C, if it is not specified which of the two segments AC and CB is to be greater than the other; but only one position, if it is specified. ​In like manner we may solve the problem, to produce a given straight line so that the square on the whole straight line made up of the given straight line and the part produced, may exceed the square on the part produced by a given square, which is not less than the square on the given straight line.
The two problems may be combined in one enunciation thus, to divide a given straight line internally or externally so that the difference of the squares ow the segments may be equal to a given square.
41. To find a point in the circumference of a given segment of a circle, so that the straight lines which join the point to the extremities of the straight line on which the segment stands may be together equal to a given straight line.
Let ACB be the circumference of the given segment, and suppose C the required point, so that the sum of AC and CB is equal to a given straight line.
Produce AC to D so that CD may be equal to CB; and join DB.
Then AD is equal to the given straight line. And the angle ACB is equal to the sum of the angles CDB and CBD (I. 32), that is, to twice the angle CDB (I. 5). Therefore the angle ADB is half of the angle in the given segment. Hence we have the following synthetical solution. Describe on AB a segment of a circle containing an angle equal to half the angle in the given segment. With A as centre, and a radius equal to the given straight line, describe a circle. Join A with a point of intersection of this circle and the segment which has been described; this ​joining straight line will cut the circumference of the given segment at a point which solves the problem.
The given straight line must exceed AB and it must not exceed a certain straight line which we will now determine. Suppose the circumference of the given segment bisected at E: join AE, and produce it to meet the circumference of the described segment at F. Then AE is equal to EB (III. 28), and EB is equal to EF for the same reason that CB is equal to CD. Thus EA, EB, EF are all equal; and therefore E is the centre of the circle of which ADB is a segment (III. 9). Hence AF is the longest straight line which can be drawn from A to the circumference of the described segment; so that the given straight line must not exceed twice AE,
42.To describe an isosceles triangle having each of the angles at the base double of the third angle.
This problem is solved in IV. 10; we may suppose the solution to have been discovered by such an analysis as the following.
Suppose the triangle ABD such a triangle as is required, so that each of the angles at B and D is double of the angle at A.
Bisect the angle at D by the straight line DC. Then the angle ADC is equal to the angle at A; therefore CA is equal to CD. The angle CBD is equal to the angle ADB, by hypothesis; the angle CDB is equal to the angle at A; therefore the third angle BCD is equal to the third angle ABD (I. 32). Therefore BD is equal to CD (I. 6); and therefore BD is equal to AC.
Since the angle BDC is equal to the angle at A, the straight line BD will touch at D the circle described round the triangle ACD {Note on III. 32). Therefore the rectangle AB, BC is equal to the square on BD (III. 36). Therefore the rectangle AB,BC is equal to the square on a AC
Therefore AB is divided at C in the manner required in II. 11.
Hence the synthetical solution of the problem is evident. ​43. To inscribe a square in a given triangle.
Let ABC be the given triangle, and suppose DEFG the required square. Draw AH perpendicular to EC, and AK parallel to EG; and let EF produced meet AK at K Then BG is to GF as BA is to AK, and EG is to GD as BA is to AH{VI. 4). But GF is equal to GD, by hypothesis. Therefore BA is to AK as BA is to AH (V. 7, V. 11). Therefore AH is equal to AK{V. 7).
Hence we have the following synthetical solution. Draw AK parallel to EG, and equal to AH; and join BK. Then BK meets AC at one of the corners of the required square, and the solution can be completed.
44. Through a given point between two given straight lines, it is required to draw a straight line, such that the rectangle contained by the parts between the given point and the given straight lines may be equal to a given rectangle.
Let P be the given point, and AE and AC the given straight lines; suppose MPN the required straight line, so that the rectangle MP, PN is equal to a given rectangle.
Produce AP to Q, so that the rectangle AP, PQ may be equal to the given rectangle. Then the rectangle MP, PN is equal to the
rectangle AP, PQ. Therefore a circle will go round AMQN {Note on III. .35). Therefore the angle PNQ is equal to the angle PAM {III. 21).
Hence we have the following synthetical solution. Produce AP to Q, so that the rectangle AP, PQ may be equal to the given rectangle; describe on PQ a segment of a circle containing an angle equal to the angle PAM; join P with a point of intersection of this circle and AC; the straight line thus drawn solves the problem. ​45. In a given circle it is required to inscribe a tri- angle so that two sides may pass through two given points, and the third side he parallel to a given straight line.
Let A and B be the given points, and CD the given straight line. Suppose PMN to be the required triangle inscribed in the given circle.
Draw NE parallel to AB join EM, and produce it if necessary to meet AB at F.
If the point F were known the problem might be considered solved. For ENM is a known angle, and therefore the chord EM is known in magnitude. And then, since F is a known point, and EM is a known magnitude, the position of M becomes known.
We have then only to shew how F is to be determined. The angle MEN is equal to the angle MFA (I. 29). The angle MEN is equal to the angle MPN (III. 21). Hence MAF and BAP are similar triangles (VI. 4). Therefore MA is to AF as BA is to AP. Therefore the rectangle MA, AP is equal to the rectangle BA, AF(VI. 16). But since J is a given point the rectangle MA, AP is known; and AB is known: thus AF is determined. ​46. In a given circle it is required to inscribe a tri- angle so that the sides may pass through three given points.
Let A, B, C be the three given points. Suppose PMN to be the required triangle inscribed in the given circle
Draw NE parallel to AB, and determine the point F as in the preceding problem. We shall then have to describe in the given circle a triangle EMN so that two of its sides may pass through given points, F and C, and the third side be parallel to a given straight line AB. This can be done by the preceding problem.
This example and the preceding are taken from the work of Catalan already cited. The present problem is sometimes called Castillon's and sometimes Cramer's; the history of the general researches to which it has given rise will be found in a series of papers in the Mathematician Vol. Ill, by the late T. S. Davies.
ON LOCI.
47. A locus consists of all the points which satisfy certain conditions and of those points alone. Thus, for example, the locus of the points which are at a given distance ​from a given point is the surface of the sphere described from the given point as centre, with the given distance as radius; for all the points on this surface, and no other points, are at the given distance from the given point. If we restrict ourselves to all the points in a fixed plane which are at a given distance from a given point, the locus is the circumference of the circle described from the given point as centre, with the given distance as radius. In future we shall restrict ourselves to loci which are situated in a fixed plane, and which are properly called plane loci.
Several of the propositions in Euclid furnish good examples of loci. Thus the locus of the vertices of all triangles which are on the same base and on the same side of it, and which have the same area, is a straight line parallel to the base; this is shewn in I. 37 and I. 39.
Again, the locus of the vertices of all triangles which are on the same base and on the same side of it, and which have the same vertical angle, is a segment of a circle described on the base; for it is shewn in III. 21, that all the points thus determined satisfy the assigned conditions, and it is easily shewn that no other points do.
We will now give some examples. In each example we ought to shew not only that all the points which we indicate as the locus do fulfil the assigned conditions, but that no other points do. This second part however we leave to the student in all the examples except the last two; in these, which are more difficult, we have given the complete investigation.
48. Required the locus of points which are equidistant from two given points.
Let A and B be the two given points; join AB; and draw a straight line through the middle point of AB at right angles to AB; then it may be easily shewn that this straight line is the required locus.
49. Required the locus of the vertices of all triangles on a given base AB, such that the square on the side terminated at A may exceed the square on the side terminated at B, by a given square.
Suppose C to denote a point on the required locus; from C draw a perpendicular on the given base, meeting it, ​produced if necessary, at D. Then the square on AC is equal to the squares on AD and CD, and the square on BC is equal to the squares on BD and CD (I. 47); therefore the. square on AC exceeds the square on BC by as much as the square on AD exceeds the square on BD. Hence D is a fixed point either in AB or in AB produced through B (40). And the required locus is the straight line drawn through D, at right angles to AB.
50. Required the locus of a point such that the straight lines drawn from it to touch two given circles may be equal
Let A be the centre of the greater circle, B the centre of a smaller circle; and let P denote any point on the required locus. Since the straight lines drawn from P to touch the given circles are equal, the squares on these straight lines are equal. But the squares on PA and PB exceed these equal squares by the squares on the radii of the respective circles. Hence the square on PA exceeds the square on PB, by a known square, namely a square equal to the excess of the square on the radius of the circle of which A is the centre over the square on the radius of the circle of which B is the centre. Hence, the required locus is a certain straight line which is at right angles to AB (49).
This straight line is called the radical axis of the two circles.
If the given circles intersect, it follows from III. 36, that the straight line which is the locus coincides with the produced parts of the common chord of the two circles.
51. Required the locus of the middle points of all the chords of a circle which pass through a fixed point.
Let A be the centre of the given circle; B the fixed ​point; let any chord of the circle be drawn so that, produced if necessary, it may pass through B. Let P be the middle point of this chord, so that P is a point on the required locus.
The straight line AP is at right angles to the chord of which P is the middle point (III. 3); therefore P is on the circumference of a circle of which AB is a diameter. Hence if B be within the given circle the locus is the circumference of the circle described on AB as diameter; if B be without the given circle the locus is that part of the circumference of the circle described on AB as diameter, which is within the given circle.
52. O is a fixed point from which any straight line is drawn meeting a fixed straight line at P; in OP a point Q is taken such that OQ is to OP in a fixed ratio: determine the locus of Q.
We shall shew that the locus of Q is a straight line.
For draw a perpendicular from O on the fixed straight line, meeting it at C; in 0C take a point D such that OD is to OC in the fixed ratio; draw from O any straight line OP meeting the fixed straight line at P, and in OP take a point Q such that OQ is to OP in the fixed ratio; join
QD. The triangles ODQ and OCP are similar (VI. 6); therefore the angle ODQ is equal to the angle OCP, and is therefore a right angle. Hence Q lies in the straight line drawn through D at right angles to OD. ​53. O is a fixed point from which any straight line is drawn meeting the circumference of a fixed circle at P ; in OP a point Q is taken such that OQ is to OP in a fixed ratio: determine the locus of Q.
We shall show that the locus is the circumfercuce of a circle.
For let C be the centre of the fixed circle; in OC take a point D such that OD is to OC in the fixed ratio, and draw any radius CP of the fixed circle; draw DQ parallel to CP meeting OP, produced if necessary, at Q. Then the triangles OCP and ODQ are similar (VI. 4), and therefore OQ is to OP as OD is to 0C, that is, in the fixed ratio. Therefore Q is a point on the locus. And DQ is to CP in the fixed ratio, so that DQ is of constant length. Hence the locus is the circumference of a circle of which D is the centre.
54. There are four given points A, B, C, D in a straight line; required the locus of a point at which AB and CD subtend equal angles.
Find a point O in the straight line, such that the rectangle OA, OD may be equal to the rectangle OB, OC (34), and take OK such that the square on OK may be equal to either of these rectangles (II. 14): the circumference of the circle described from as centre, with radius OK shall be the required locus.
[We will take the case in which the points are in the following order, O, A, B, C, D.]
For let P be any point on tho circumference of this ​circle. Describe a circle round PAD, and also a circle
round PBC; then OP touches each of these circles (III. 37); therefore the angle OPA is equal to the angle PDA, and the angle OPB is equal to the angle PCB {III. 32). But the angle OPB is equal to the angles OPA and APB together, and the angle PCB is equal to the angles CPD and PDA together (I. 32). Therefore the angles OPA and APB together are equal to the angles CPD and PDA together; and the angle OPA has been shewn equal to the angle PDA; therefore the angle APB is equal to the angle CPD.
We have thus shewn that any point on the circumference of the circle satisfies the assigned conditions; we shall now shew that any point which satisfies the assigned conditions is on the circumference of the circle.
For take any point Q which satisfies the required conditions. Describe a circle round QAD, and also a circle round QBC. These circles will touch the same straight line at Q; for the angles AQB and CQD are equal, and the converse of III. 32 is true. Let this straight line which touches both circles at Q be drawn; and let it meet the straight line containing the four given points at R. Then the rectangle RA, RD is equal to the rectangle RB,RC; for each is equal to the square on RQ (III. 36). Therefore R must coincide with O (34); and therefore RQ must be equal to OK. Thus Q must be on the circumference of the circle of which is the centre, and OK the radius.
​55. Required the locus of the vertices of all the triangles ABC which stand on a given base AB, and have the side AC to the side BC in a constant ratio.
If the sides AC and BC are to be equal, the locus is
the straight lino which bisects AB at right angles. "We will suppose that the ratio is greater than a ratio of equality; so that AC is to be the greater side.
Divide AB at D so that AD is to DB in the given ratio (VI. 10); and produce AB to E, so that AE is to EB in the given ratio. Let P be any point in the required locus; join PD and PE. Then PD bisects the angle APB, and PE bisects the angle between BP and AP produced. Therefore the angle DPE is a right angle. Therefore P is on the circumference of a circle described on DE as diameter.
We have thus shewn that any point which satisfies the assigned conditions is on the circumference of the circle described on DE as diameter; we shall now shew that any point on the circumference of this circle satisfies the assigned conditions.
Let Q be any point on the circumference of this circle, QA shall be to QB in the assigned ratio. For, take O the centre of the circle; and join QO. Then, by construction, AE to EB as AD is to DB, and therefore, alternately, AE to AD as EB is to DB; therefore the sum of AE and AD is to their difference as the sum of EB and DB is to their difference (23); that is, twice DO is to twice DO as twice DO is to twice BO; therefore AO is to DO as DO is ​to BO that is, AO is to OQ as QO is to OB. Therefore the triangles AOQ and QOB are similar triangles (VI. 6); and therefore AQ is to QB as QO is to BO. This shews that the ratio of AQ to BQ is constant; we have still to shew that this ratio is the same as the assigned ratio.
We have already shewn that AO is to DO as DO is to BO; therefore, the difference of AO and BO is to DO as the difference of DO and BO is to BO (V. 17); that is, AD is to DO as BD is to BO; therefore AD is to BD as DO is to BO; that is, AD is to DB as QO is to BO. This shews that the ratio of QO to BO is the same as the assigned ratio.
ON MODERN GEOMETRY.
56. We have hitherto restricted ourselves' to Euclid's Elements, and propositions which can be demonstrated by strict adherence to Euclid's methods. In modern times various other methods have been introduced, and have led to numerous and important results. These methods may be called semi-geometrical, as they are not confined within the limits of the ancient pure geometry; in fact the power of the modern methods is obtained chiefly by combining arithmetic and algebra with geometry. The student who desires to cultivate this part of mathematics may consult Townsend's Chapters on the Modern Geometry of the Pointy Line, and Circle.
We will give as specimens some important theorems, taken from what is called the theory of transversals.
Any line, straight or curved, which cuts a system of other lines is called a transversal; in the examples which we shall give, the lines will be straight lines, and the system svill consist of three straight lines forming a triangle.
We will give a brief enunciation of the theorem which we are about to prove, for the sake of assisting the memory in retaining the result; but the enunciation will not be fully comprehended until the demonstration is completed. ​57. If a straight line cut the sides, or the sides produced, of a triangle, the product of three segments in order is equal to the product of the other three segments.
Let ABC be a triangle, and let a straight line be drawn cutting the side BC at D, the side CA at E, and the side AB produced through B at F. Then BD and DC are
called segments of the side BC, and CE and EA are called segments of the side CA, and also AF and FB are called segments of the side AB.
Through A draw a straight line parallel to BC, meeting DF produced at H.
Then the triangles CED and EAH are equiangular to one another; therefore AH is to CD as AE is to EC (VI. 4). Therefore the rectangle AH, EC is equal to the rectangle CD, AE (VL 16).
Again, the triangles FAH and FBD are equiangular to one another; therefore AH is to BD as FA is to FB (VI. 4). Therefore the rectangle AH, FB is equal to the rectangle BD, FA (VI. 16).
Now suppose the straight lines represented by numbers in the manner explained in the notes to the second Book of the Elements. We have then two results which we can express arithmetically: namely, the prooduct AH.EC is equal to the product CD.AE; and the product AH.FB is equal to the product BD.FA.
Therefore, by the principles of arithmetic, the product AH.EC.BD.FA is equal to the product AH.FB.CD.AE, and therefore, by the principles of arithmetic, the product BD.CE.AF is equal to the product DC.EA. FB.
This is the result intended by the enunciation given above. Each product is made by three segments, one from ​every side of the triangle: and the two segments which terminated at any angular point of the triangle are never in the same product. Thus if wo begin one product with the segment BD, the other segment of the side BC, namely DC, occurs in the other product; then the segment CE occurs in the first product, so that the two segments CD and CE, which terminate at C, do not occur in the same product; and so on.
The student should for exercise draw another figure for the case in which the transversal meets all the sides produced, and obtain the same result.
58. Conversely, it may be shewn by an indirect proof that if the product BD.CE.AF be equal to the product DC.EA.FB, the three points D,E,F lie in the same straight line.
59. If three straight lines he drawn through the angular points of a triangle to the opposite sides, and meet at the same pointy the product of three segments in order is equal to the product of the other three segments.
Let ABC be a triangle. From the angular points to the opposite sides let the straight lines AOD, BOE, COF be drawn, which meet at the point O: the product AF.BD.CE shall be equal to the product FB.DC.EA.
For the triangle ABD is cut by the transversal FOC, and therefore by the theorem in 57 the following products are equal, AF.'BC.DO, and FB.CD.OA.
Again, the triangle ACD is cut by the transversal EOB, and therefore by the theorem in 57 the following products are equal, AO.DB.CE and OD.BC.EA.
​Therefore, by the principles of arithmetic, the following products are equal, AF.BC.DO.AO.DB.CE and FB.CD.OA.OD.BC.EA. Therefore the following
products are equal, AF.BD.GE and FB.DC. EA. W have supposed the point to be within the triangle; if O be without the triangle two of the points D, E, F will fall on the sides produced.
60. Conversely, it may be shewn by an indirect proof that if the product AF. BD . CE be equal to the product FB.DC.EA, the three straight lines AD, BE, CF meet at the same point.
6l. We may remark that in geometrical problems the following terms sometimes occur, used in the same sense as in arithmetic; namely arithmetical progression, geometrical progression, and harmonical progression. A proposition respecting harmonical progression, which deserves notice, will now be given.
62. Let ABC he a triangle; let the angle A be bisected by a straight line which meets BC at D, and let the exterior angle at A be bisected by a straight line which meets BC produced through C, at E: then BD, BC, BE shall he in harmonical progression.
​For BD is to DC as BA is to AC (VI. 3); and BE is to AC as BA is to AC (VI. A). Therefore BD is to DC as BE is to EC (V. 11). Therefore BD is to BE as DC is to EC (V. 16). Thus of the three straight lines BD,BC, BE, the first is to the third as the excess of the second over the first is to the excess of the third over the second. Therefore BD, BC, BE are in harmonical progression.
This result is sometimes expressed by saying that BE is divided harmonically at D and C.