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

BOOK IV.
DEFINITIONS.
1. A rectilneal figure is said to be inscribed in another rectilineal figure, when all the angles of the inscribed figure are on the sides of the figure in which it is inscribed, each on each.
2. In like manner, a figure is said to be described about another figure, when all the sides of the circumscribed figure pass through the angular points of the figure about which it is described, each through each.
3. A rectilineal figure is said to be inscribed in a circle, when all the angles of the inscribed figure are on the circumference of the circle.
4. A rectilineal figure is said to be described about a circle, When each side of the circumscribed figure touches the circumference of the circle.
5. In like manner, a circle is said to be inscribed in a rectilineal figure, when the circumference of the circle ouches each side of the figure. ​ 6. A circle is said to be described about a rectilineal figure, when the circircumference of the circle passes through all the angular points of the figure about which it is described.
7. A straight line is said to be placed in a circle, when the extremities of it are in the circumference of the circle.
PROPOSITION 1. PROBLEM.
In a given circle, to place a straight line, equal to a given straight line, which is not greater than the diameter of the circle.
Let ABC be the given circle, and D the given straight line, not greater than the diameter of the circle: it is required to place in the circle ABC, a straight line equal to D.
Draw BC, a diameter of the circle ABC. Then, if BC is equal to D, the thing required is done; for in the circle ABC, a straight line is placed equal to D.
But, if it is not, BC is greater than D. [Hypothesis. Make CE equal to D, and from the centre C, at the distance CE, describe the circle AEF, and join CA.
Then, because C is the centre of the circle AEF, CA is equal to CE; [I. Definition 15. but CE is equal to D; [Construction. therefore CA is equal to D. [Axiom 1.
PROPOSITION 2. PROBLEM.
In a given circle, to inscribe a triangle equiangular to a given triangle.
Let ABC be the given circle, and DEF the given triangle: it is required to inscribe in the circle ABC a triangle equiangular to the triangle DEF.
Draw the straight line GAH touching the circle at the point A; [III. 17. at the point A, in the straight line AH, make the angle GAB equal to the angle HAC;[I 23. and, at the point A, in the straight line AG, make the angle GAB equal to the angle DFE; and join BC. ABC shall be the triangle required.
Because GAH touches the circle ABC, and AC is drawn from the point of contact A, [Construction. therefore the angle HAC is equal to the angle ABC in the alternate segment of the circle. [III. 32. But the angle HAC is equal to the angle DEF. [Constr. Therefore the angle ABC is equal to the angle DEF. [Ax.l.
For the same reason the angle ACB is equal to the angle DFE. Therefore the remaining angle BAC is equal to the remaining angle EDF. [I. 32, Axioms 11 and 3.
Wherefore the triangle ABC is equiangular to the triangle DEF, and it is inscribed in the circle ABC. q.e.f.
PROPOSITION 3. PROBLEM.
About a given circle, to describe a triangle equiangular to a given triangle. ​Let ABC be the given circle, and DEF the given triangle: it is required to describe a triangle about the circle ABC, equiangular to the triangle DEF.
Produce EF both ways to the points G, H; take K the centre of the circle ABC; [III. 1.
from K draw any radius KB; at the point K, in the straight line KB, make the angle BKA equal to the angle DEG, and the angle BKC equal to the angle DFH; [I. 23. and through the points A, B, C, draw the straight lines LAM, MBN, NCL, touching the circle ABC. [III. 17. LMN shall be the triangle required.
Because LM, MN, NL touch the circle ABC at the points A, B, C, [Construction.
to which from the centre are drawn KA, KB, KC, therefore the angles at the points A,B,C are right angles.[III.18. And because the four angles of the quadrilateral figure AMBK are together equal to four right angles,/br> for it can be divided into two triangles, and that two of them KAM, KBM are right angles, therefore the other two AKB, AMB are together equal to two right angles. [Axiom. 3. But the angles DEG,DEF are together equal to two right angles. [I. 13.
Therefore the angles AKB, AMB are equal to the angles DEG, DEF; of which the angle AKB is equal to the angle DEG; [Constr.
therefore the remaining angle AMB is equal to the remaining angle DEF. [Axiom 3.
In the same manner the angle LNM may be shewn to be equal to the angle DFE.
Therefore the remaining angle MLN is equal to the remaining angle EDF. [I. 32, Axioms 11 and 3. ​Wherefore the triangle LMN is equiangular to the triangle DEF, and it is described about the circle ABC. q.e.f.
PROPOSITION 4. PROBLEM.
To inscribe a circle in a given triangle.
Let ABC be the given triangle: it is required to inscribe a circle in the triangle ABC
Bisect the angles ABC, ACB, by the straight lines BD, CD meeting one another at the point D; [I.9 and from D draw DE, DF, DG perpendiculars to AB, BC, CA. [1.12. Then, because the angle EBD IS equal to the angle FBD, for the angle ABC is bisected by BD [Construction. and that the right angle BED is equal to the right angle BFD; [Axiom 11. therefore the two triangles EBD, FBD have two angles and the side BD, which is opposite to one of the equal angles in each, is common to both; therefore DE is equal to DF.
For the same reason DG is equal to DF. Therefore DE is equal to DG. [Axiom 1.
Therefore the three straight lines DE, DF, DG are equal to one another, and the circle described from the centre D, at the distance of any one of them, will pass through the extremities of the other two; and it will touch the straight lines AB, BC, CA because the angles at the points E, F, G are right angles, and the straight line which is drawn from the extremity of a dia- meter at right angles to it, touches the circle. [III. 16. Cor. Therefore the straight line AB,BC,CA do each of them touch the circle, and therefor the circle is inscribed in the triangle ABC.
PROPOSITION 5. PROBLEM.
To describe a circle about a given triangle.
Let ABC be the given triangle: it is required to describe a circle about ABC.
Bisect AB, AC at the points D, E; [I. 10. from these points draw DF, EF, at right angles to AB,AC; [I. 11. DF, EF, produced, will meet one another; for if they do not meet they are parallel, therefore AB, AC, which are at right angles to them are parallel; which is absurd: let them meet at F, and join FA; also if the point F be not in BC, join BF, CF.
Then, because AD is equal to BD [Construction. and DF is common, and at right angles to AB, therefore the base FA is equal to the base FB. [I. 4. In the same manner it may be shewn that FC is equal to FA. Therefore FB is equal to FC; [Axiom 1. and FA, FB, FC are equal to one another. Therefore the circle described from the centre F, at the distance of any one of them, will pass through the extremities of the other two, and will be described about the triangle ABC.
Wherefore a circle has been described about the given triangle. q.e.f.
Corollary. And it is manifest, that when the centre of the circle falls within the triangle, each of its angles is less than a right angle, each of them being in a segment greater than a semicircle; and when the centre is in one of the sides of the triangle, the angle opposite to this side, being in a semicircle, is a right angle; and when the centre ​falls without the triangle, the angle opposite to the side beyond which it is, being in a segment less than a semicircle, is greater than a right angle. [III. 31. Therefore, conversely, if the given triangle be acuteangled, the centre of the circle falls within it; if it be a right-angled triangle, the centre is in the side opposite to the right angle; and if it be an obtuse-angled triangle, the centre falls without the triangle, beyond the side opposite to the obtuse angle.
PROPOSITION 6. PROBLEM.
To inscribe a square in a given circle.
Let ABCD be the given circle: it is required to inscribe a square in ABCD.
Draw two diameters AC, BD of the circle ABCD, at right angles to one another; [III. 1, I. 11. and join AB, BC, CD, DA. The figure ABCD shall be the square required.
Because BE is equal to DE, for E is the centre; and that EA is common, and at right angles to BD;
therefore the base BA is equal to the base DA. [I. 4. And for the same reason BC, DC are each of them equal to BA, or DA. Therefore the quadrilateral figure ABCD is equilateral.
It is also rectangular.
For the straight line BD being a diameter of the circle ABCD, BAD is a semicircle; [Construction. therefore the angle BAD is a right angle. [III. 31. For the same reason each of the angles ABC, BCD, CDA is a right angle;
therefore the quadrilateral figure ABCD is rectangular.
And it has been shewn to be equilateral; therefore it is a square.
PROPOSITION 7. PROBLEM.
To describe a square about a given circle.
Let ABCD be the given circle: it is required to describe a square about it.
Draw two diameters AC, BD of the circle ABCD, at right angles to one another; [III. 1, 1. 11. and through the points A,B,C,D, draw FG, GH, HK, KF touching the circle. [III. 17. The figure GHKF shall be the square required.
Because FG touches the circle ABCD, and EA is drawn from the centre E to the point of contact A, [Construction. therefore the angles at A are right angles. [III. 18. For the same reason the angles at the points B, C, D are right angles.
And because the angle AEB is a right angle, [Construction. and also the angle EBG is a right angle, therefore GH is parallel to AC. [I. 28.
For the same reason AC is parallel to FK.
In the same manner it may be shewn that each of the lines GF, HK is parallel to BD.
Therefore the figures GK, GC, CF, FB, BK are parallelograms; and therefore GF is equal to HK, and GH to FK. [I. 34.
And because AC is equal to BD, and that AC is, equal to each of the two GH, FK, and that BD is equal to each of the two GF, HK, therefore GH, FK are each of them equal to GF, or HK; therefore the quadrilateral figure FGHK is equilateral.
It is also rectangular. For since AEBG is a parallelogram, and AEB a right angle, therefore AGB is also a right angle. [I. 34. In the same manner it may be shewn that the angles at H, K, F are right angles; ​therefore the quadrilateral figure FGHK is rectangular.
And it has been shewn to be equilateral; therefore it is a square.
Wherefore a square has been described about the given circle. q.e.f.
PROPOSITION 8. PROBLEM.
To inscribe a circle in a given square.
Let ABCD be the given square: it is required to inscribe a circle in ABCD.
Bisect each of the sides AB, AD at the points F, E; [I. 10. through E draw EH parallel to AB or DC, and through F draw FK parallel to AD or BC. [I. 31.
Then each of the figures AK, KB, AH, HD, AG, GC, BG, GD is a right-angled parallelogram; and their opposite sides are equal. [I. 34. And because AD is equal to AB, [I. Definition 30. and that AE is half of AD, and AF half of AB, [Constr. therefore AE is, equal to AF. [Axiom 7.
Therefore the sides opposite to these are equal, namely, FG equal to GE. [I. 34.
In the same manner it may be shewn that the straight lines GH, GK are each of them equal to FG or GE. Therefore the four straight lines GE, GF, GH, GK are equal to one another, and the circle described from the centre G, at the distance of any one of them, will pass through the extremities of the other three; and it will touch the straight lines AB, BC, CD, DA, because the angles at the points E, F, H, K are right angles, and the straight line which is drawn from the extremity of a diameter, at right angles to it, touches the circle. [III. 16. Corollary.
Therefore the straight lines AB, BC, CD, DA do each of them touch the circle.
PROPOSITION 9. PROBLEM.
To describe a circle about a given square.
Let ABCD be the given square: it is required to describe a circle about ABCD.
Join AC, BD, cutting one another at E.
Then, because AB is equal to AD,
and AC is common to the two triangles BAC, DAC; the two sides BA, AC are equal to the two sides DA,AC each to each; and the base BC is equal to the base DC; therefore the angle BAC is equal to the angle DAC, [I. 8. and the angle BAD is bisected by the straight line AC.
In the same manner it may be shewn that the angles ABC, BCD, CDA are severally bisected by the straight lines BD, AC.
Then, because the angle DAB is equal to the angle ABC, and that the angle EAB is half the angle DAB, and the angle EBA is half the angle ABC, therefore the angle EAB is equal to the angle EBA; [Ax. 7. and therefore the side EA is equal to the side EB. [I. 6.
In the same manner it may be shewn that the straight lines EC, ED are each of them equal to EA or EB. Wherefore the four straight lines EA, EB, EC, ED are equal to one another, and the circle described from the centre E, at the distance of any one of them, will pass through the extremities of the other three, and will be described about the square ABCD.
Wherefore a circle has been described about the given sqaure. q.e.f.
PROPOSITION 10. PROBLEM.
To describe an isosceles triangle, having each of the angles at the base double of the third angle. ​ Take any straight line AB and divide it at the point C, so that the rectangle AB, BC may be equal to the square on AC; [II. 11. from the centre A, at the distance AB describe the circle BDE, in which place the straight line BD equal to AC, which is not greater than the diameter of the circle BDE; [IV. 1. and join DA. The triangle ABD shall be such as is required; that is, each of the angles ABD, ADB shall be double of the third angle BAD.
Join DC; and about the triangle ACD describe the circle ACD. [IV. 5. Then, because the rectangle AB, BC is equal to the square on AC, [Construction. and that AC is equal to BD, [Construction. therefore the rectangle AB, BC is equal to the square on BD.
And, because from the point B, without the circle ACD, two straight lines BCA, BD are drawn to the circumference, one of which cuts the circle, and the other meets it, and that the rectangle AB, BC, contained by the whole of the cutting line, and the part of it without the circle, is equal to the square on BD which meets it; therefore the straight line BD touches the circle. [I II. 37.
And, because BD touches the circle ACD, and DC is drawn from the point of contact D, therefore the angle BDC is equal to the angle DAC in the alternate segment of the circle. [III. 32. To each of these add the angle CDA; therefore the whole angle BDA is equal to the two angles CDA, DAC. [Axiom 2. But the exterior angle BCD is equal to the angles CDA DAC. [1.32. ​ Therefore the angle BDA is equal to the angle BCD [[Ax .l. But the angle BDA is equal to the angle DBA, [I. 5. because AD is equal to AB. Therefore each of the angles BDA, DBA, is equal to the angle BCD. [Axiom 6.
And, because the angle DBC is equal to the angle BCD, the side DB is equal to the side DC; [1.6. but DB was made equal to CA; therefore CA is equal to CD, [Axiom 6. and therefore the angle CAD is equal to the angle CDA. [I.5 Therefore the angles CAD, CDA are together double the angle CAD.
But the angle BCD is equal to the angles CAD, CDA. [I.32 Therefore the angle BCD is double of the angle CAD, And the angle BCD has been shewn to be equal to each of the angles BDA, DBA; therefore each of the angles BDA, DBA is double of the angle BAD.
Wherefore an isosceles triangle has been described, having each of the angles at the base double of the third angle, q.e.f.
PROPOSITION 11. PROBLEM.
To inscribe an equilateral and equiangular pentagon in a given circle.
Let ABCDE be the given circle: it is required to inscribe an equilateral and equiangular pentagon in the circle ABCDE.
Describe an isosceles triangle, FGH, having each of the angles at G, H, double of the angle at F [IV. 10. in the circle ABCDE, inscribe the triangle ACD, equiangular to the triangle FGH, so that the angle CAD may ​ be equal to the angle at F, and each of the angles ACD, equal to the angle at G or H [IV. 2. and therefore each of angles ACD,ADC is double of the angle CAD; bisect the angles ACD, ADC by the straight line CE,DB; [1.9. and join AB, BC, AE, ED. ABCDE shall be the pentagon required.
For because each of the angles ACD, ADC is double of the angle CAD, and that they are bisected by the straight lines CE, DB, therefore the five angles ADB, BDC, CAD, DCE, ECA equal to one another. equal angles stand on equal arcs; [III. 26. therefore the five arcs AB, BC, CD, DE, EA are equal to one another. And equal arcs are subtended by equal straight lines; [III. 29.
therefore the five straight lines AB, BC, CD, DE, EA are equal to one another; and therefore the pentagon ABCDE is equilateral.
It is also equiangular. For, the arc AB is equal to the arc DE; each of these add the arc BCD; therefore the whole arc ABCD is equal to the whole arc BCDE. [Axiom 2.
And the angle AED stands on the arc ABCD, and the angle BAE on the arc BCDE. Therefore the angle AED is equal to the angle BAE. [III. 27. For the same reason each of the angles ABC, BCD, DE is equal to the angle AED or BAE; Therefore the pentagon ABCDE is equiangular. And it has been shewn to be equilateral.
PROPOSITION 12. PROBLEM.
To describe an equilateral and equiangular pentagon about a given circle.
Let ABCDE be the given circle: it is required to describe an equilateral and equiangular pentagon about the circle ABCDE.
Let the angles of a pentagon, inscribed in the circle, by the last proposition, be at the points A, B, C, D, E, so that the arcs AB, BC, CD, DE, EA are equal; and through the points A, B, C, D, E, draw GH, HK, KL, LM, MG touching the circle. [III 17. The figure GHKLM shall be the pentagon required.
Take the centre F, and join FB, FK, FC, FL, FD. Then, because the straight line KL touches the circle ABCDE at the point C to which FC is drawn from the centre, therefore FC is perpendicular to KL [III. 18. therefore each of the angles at C is a right angle. For the same reason the angles at the points B, D are right angles.
And because the angle FCK is, a right angle, the square on FK equal to the squares on FC, CK. [I. 47. For the same reason the square on FK is equal to the squares on FB, BK.
Therefore the squares on FC, CK are equal to the squares on FB,BK; [Axiom 1. of which the square on FC is equal to the square on FB; therefore the remaining square on CK is equal to the remaining square on BK, [Axiom 3. and therefore the straight line CK is equal to the straight line BK. ​And because FB is equal to FC,
and FK is common to the two triangles BFK, CFK ;
the two sides BF, FK are equal to the two sides CF, FK, each to each ;
and the base BK was shewn equal to the base CK;
therefore the angle BFK is equal to the angle CFK, [I. 8.
and the angle BKF to the angle CKF. [I. 4.
Therefore the angle BFC is double of the angle CFK, and the angle BKC is double of the angle CKF.
For the same reason the angle CFD is double of the angle CFL, and the angle CLD is double of the angle CLF.
And because the arc BC is equal to the arc CD, the angle BFC is equal to the angle CFD ; [III. 27.
and the angle BFC is double of the angle CFK, and the angle CFD is double of the angle CFL ;
therefore the angle CFK is equal to the angle CFL. [Ax. 7.
And the right angle FCK is equal to the right angle FCL.
Therefore in the two triangles FCK, FCL, there are two angles of the one equal to two angles of the other, each to each ;
and the side FC, which is adjacent to the equal angles in each, is common to both ;
therefore their other sides are equal, each to each, and the third angle of the one equal to the third angle of the other ;
therefore the straight line CK is equal to the straight line CL, and the angle FKC to the angle FLC. [I. 26.
And because CK is equal to CL, LK is double of CK.
In the same manner it may be shewn that HK is double of BK.
And because BK is equal to CK, as was shewn, and that HK is double of BK, and LK double of CK, therefore HK is equal to LK. [Axiom 6.
In the same manner it may be shewn that GH, txM, ML are each of them equal to HK or LK; therefore the pentagon GHKLM is equilateral.
It is also equiangular. ​ For, since the angle FKC is equal to the angle FLC, and that the angle HKL is double of the angle FKC, and the angle KLM double of the angle FLC, as was shewn,
therefore the angle HKL is equal to the angle KLM. [Axiom 6.
In the same manner it may be shewn that each of the angles KHG, HGM, GML is equal to the angle HKL or KLM; therefore the pentagon GHKLM is equiangular. And it has been shewn to be equilateral.
Wherefore an equilateral and equiangular pentagon has been described about the given circle. q.e.f.
PROPOSITION 13. PROBLEM.
To inscribe a circle in a given equilateral and equiangular pentagon.
Let ABCDE be the given equilateral and equiangular pentagon: it is required to inscribe a circle in the pentagon ABCDE.
Bisect the angles BCD, CDE by the straight lines CF,DF; [1.9. and from the point F, at which they meet, draw the straight lines FB, FA, FE.
Then, because BC is equal to DC [Hypothesis.
and CF is common to the two triangles BCF, DCF; the two sides BC, CF are equal to the two sides DC, CF, each to each;
and the angle BCF is equal to the angle DCF; [Constr. therefore the base BF is equal to the base DF, and the ​other angles to the other angles to which the equal sides are opposite ; [I. 4.
therefore the angle CBF is equal to the angle CDF.
And because the angle CDE is double of the angle CDF, and that the angle CDE is equal to the angle CBA, and the angle CDF is equal to the angle CBF,
therefore the angle CBA is double of the angle CBF;
therefore the angle. ABF is equal to the angle CBF;
therefore the angle ABC is bisected by the straight line BF.
In the same manner it may be shewn that the angles BAE, AED are bisected by the straight lines AF, EF.
From the point F draw FG, FH, FK, FL, FM perpen- diculars to the straight lines AB, BC, CD, DE, EA. [1. 12.
Then, because the angle FCH is equal to the angle FCK, and the right angle FHC equal to the right angle FKC;
therefore in the two triangles FHC, FKC, there are two triangles of the one equal to two angles of the other, each to each; and the side FC, which is opposite to one of the equal angles in each, is common to both ; therefore their other sides are equal, each to each, and therefore the perpendicular FH is equal to the perpen- dicular FK. [I. 26.
In the same manner it may be shewn that FL, FM, FG are each of them equal to FH or FK.
Therefore the five straight lines FG, FH, FK, FL, FM are equal to one another, and the circle described from the centre F, at the distance of any one of them will pass through the extremities of the other four ; and it will touch the straight lines AB, BC, CD, DE, EA, because the angles at the points G, H, K, L, M are right angles, [Construction.
and the straight line drawn from the extremity of a dia- " meter, at right angles to it, touches the circle ; [III. 16.
Therefore each of the straight lines AB, BC, CD, DE, EA touches the circle.
PROPOSITION 14. PROBLEM
To describe a circle about a given equilateral and equiangular pentagon.
Let ABCDE be the given equilateral and equiangular pentagon: it is required to describe a circle about it.
Bisect the angles BCD, CDE by the straight lines CF, DF; [1.9. and from the point F, at which they meet, draw the straight lines FB, FA, FE.
Then it may be shewn, as in the preceding proposition, that the angles CBA, BAE, AED are bisected by the straight lines BF, AF, EF.
And, because the angle BCD is equal to the angle CDE, and that the angle FCD is half of the angle BCD, and the angle FDC is half of the angle CDE, therefore the angle FCD is equal to the angle FDC, [Ax. 7.
therefore the side FC is equal to the side FD. [I. 6.
In the same manner it may be shewn that FB, FA, FE are each of them equal to FC or FD; therefore the five straight lines FA, FB, FC, FD, FE are equal to one another, and the circle described from the centre F, at the distance of any one of them, will pass through the extremities of the other four, and will bo described about the equilateral and equiangular pentagon ABCDE.
Wherefore a circle has been described about the given equilateral and equiangular pentagon. q.e.f.
PROPOSITION 15. PROBLEM.
To inscribe an equilateral and equiangular hexagon in a given circle
Let ABCDEF be the given circle: it is requirerd to inscribe an equilateral and equiangular hexagon in it. Find the centre G of the circle ABCDEF, ​ and draw the diameter AGD; from the centre D, at the distance DG dscribe the circle EGCH; join EG, CG, and produce them to the points B,F; and join AB, BC, CD, DB, BF, FA. The hexagon ABCDBF shall be equilateral and equiangular.
For, because G is the centre of the circle ABCDBF, GE is equal to GD; and because D is the centre of the circle EGCH, DB is equal to DG; therefore GE is equal to DE,[Axiom 1. and the triangle EGD is equilateral;
therefore the three angles EGD, GDE, DEG are equal to one another. [I. 5. Corollary. But the three angles of a triangle are together equal to two right angles; [I. 32. therefore the angle EGD is the third part of two right angles.
In the same manner it may be shewn, that the angle DGC is the third part of two right angles.
And because the straight line GC makes with the straight line EB the adjacent angles EGC, CGB together equal to two right angles, [I. 13. therefore the remaining angle CGB is the third part of two right angles; therefore the angles EGD, DGC, CGB are equal to one aother.
And to these are equal the vertical opposite angles BGA,AGF,FGB. [1.15. therefore the six angles EGD, DGC, CGB, BGA, AGF, FGE are equal to one another. ​But equal angles stand on equal arcs; [III. 26. therefore the six arcs AB, BC, CD, DE, EF, FA are equal to one another. And equal arcs are subtended by equal straight lines; [III.29. therefore the six straight lines are equal to one another, and the hexagon is equilateral.
It is also equiangular. For, the arc AF is equal to the arc ED; to each of these add the arc ABCD; therefore the whole arc FABCD is equal to the whole arc ABCDE; and the angle FED stands on the arc FABCD, and the angle FED stands on the arc ABCDE; therefore the angle FED is equal to the angle AFE. [III. 27.
In the same manner it may be shewn that the other angles of the hexagon ABCDEF are each of them equal to the angle AFE or FED; therefore the hexagon is equiangular. And it has been shewn to be equilateral; and it is inscribed in the circle ABCDEF.
Wherefore an equilateral and equiangular hexagon has been inscribed in the given circle, q.e.f.
Corollary. From this it is manifest that the side of the hexagon is equal to the straight line from the centre, that is, to the semidiameter of the circle.
PROPOSITION 16. PROBLEM.
To inscribe an equilateral and equiangular quindecagon in a given circle.
Let ABCD be the given circle: it is required to inscribe an equilateral and equiangular quindecagon in the circle ABCD.
Let AC be the side of an equilateral triangle inscribed in the circle; [IV. 2. and let AB be the side of an equilateral and equiangular pentagon inscribed in the circle. [IV. 11.
Then, of such equal parts as the whole circumference ABCDEF contains fifteen, the arc ABC, which is the third part of the whole, contains five, and the arc AB, which is tht; fifth part of the whole, contains three; therefore their difference, the arc BC, contains two of the same parts. Bisect the arc BC at E; [III. 30. therefore each of the arcs BE, EC is the fifteenth part of the whole circumference ABCDF.
Therefore if the straight lines BE, EC be drawn, and straight lines equal to them be placed round in the whole circle, [IV. 1. an equilateral and equiangular quindecagon will be inscribed in it. q.e.f.
And, in the same manner as was done for the pentagon, if through the points of division made by inscribing the quindecagon, straight lines be drawn touching the circle, an equilateral and equiangular quindecagon will be described about it; and also, as for the pentagon, a circle may be inscribed in a given equilateral and equiangular quindecagon, and circumscribed about it.