Geometry

The ***Elements*** ( *Stoicheia*) is a mathematical treatise consisting of 13 books attributed to the ancient Greek mathematician Euclid in Alexandria, Ptolemaic Egypt c. 300 BC. It is a collection of definitions, postulates, propositions (theorems and constructions), and mathematical proofs of the propositions. The books cover plane and solid Euclidean geometry, elementary number theory, and incommensurable lines. *Elements* is the oldest extant large-scale deductive treatment of mathematics. It has proven instrumental in the development of logic and modern science, and its logical rigor was not surpassed until the 19th century.

Euclid's *Elements* has been referred to as the most successful and influential textbook ever written. It was one of the very earliest mathematical works to be printed after the invention of the printing press and has been estimated to be second only to the Bible in the number of editions published since the first printing in 1482, with the number reaching well over one thousand. For centuries, when the quadrivium was included in the curriculum of all university students, knowledge of at least part of Euclid's *Elements* was required of all students. Not until the 20th century, by which time its content was universally taught through other school textbooks, did it cease to be considered something all educated people had read.

## Let it be granted

1. That a straight line may be drawn from any one point to any other point:

2. That a terminated straight line may be produced to any length in a straight line:

3. And that a circle may be described from any centre, at any distance from that centre.

## AXIOMS

1. Things which are equal to the same thing are equal to one another.

2. If equals be added to equals the wholes are equal.

3. If equals be taken from equals the remainders are equal.

4. If equals be added to unequals the wholes are unequal.

5. If equals be taken from unequals the remainders are unequal.

6. Things which are double of the same thing are equal to one another.

7. Things which are halves of the same thing are equal to one another.

8. Magnitudes which coincide with one another, that is, which exactly fill the same space, are equal to one another.

9. The whole is greater than its part.

10. Two straight lines cannot enclose a space.

11. All right angles are equal to one another.

12. If a straight line meet two straight lines, so as to make the two interior angles on the same side of it taken together less than two right angles, these straight lines, being continually produced, shall at length meet on that side on which are the angles which are less than two right angles.

## PROPOSITION 1. *PROBLEM.*

To describe an equilateral triangle on a given finite straight line.

Let *AB* be the given straight line; it is required to describe an equilateral triangle on *AB*.

From the centre *A* at the distance *AB* describe the circle *BCD*. [*Postulate* 3.

From the centre *B*, at the distance *BA*, describe the circle *ACE*. [*Postulate* 3.

From the point *C*, at which the circles cut one another, draw the straight lines *CA* and *CB* to the points *A* and *B*. [*Post*. 1.

*ABC* shall be an equilateral triangle.

Because the point *A* is the centre of the circle *BCD*, *AC* is equal to *AB*. [*Definition* 15.

And because the point *B* is the centre of the circle *ACE*, *BC* is equal to *BA*. [*Definition* 15.

But it has been shewn that *CA* is equal to *AB*;

 therefore *CA* and *CB* are each of them equal to *AB*.

But things which are equal to the same thing are equal to one another. [*Axiom* 1.

Therefore *CA* is equal to *CB*.

Therefore *CA*, *AB*, *BC* are equal to one another.

Wherefore the *triangle ABC is equilateral*, [*Def*. 24. *and it is described on the given straight line AB*. Q.E.F.

## PROPOSITION 2. *PROBLEM.*

From a given point to draw a straight line equal to a given straight line.

Let *A* be the given point, and *BC* the given straight line: it is required to draw from the point *A* a straight line equal to *BC*.

From the point *A* to *B* draw the straight line *AB*; [*Post*. 1.

and on it describe the equilateral triangle *DAB*, [I. 1.

and produce the straight lines *DA*, *DB* to *E* and *F*. [*Post*. 2.

From the centre *B*, at the distance *BC*, describe the circle *CGH*, meeting *DF* at *G*. [*Post*. 3.

From the centre *D*, at the distance *DG*, describe the circle *GKL*, meeting *DE* at *L.* [*Post*. 3.

*AL* shall be equal to *BC*.

Because the point *B* is the centre of the circle *CGH*, *BC* is equal to *BG*. [*Definition* 15.

And because the point *D* is the centre of the circle *GKL*, *DL* is equal to *DG*; [*Definition* 15.

and *DA*, *DB* parts of them are equal; [*Definition* 24.

therefore the remainder *AL* is equal to the remainder BG. [*Axiom* 3.

But it has been shewn that *BC* is equal to *BG*;

therefore *AL* and *BC* are each of them equal to *BG*.

But things which are equal to the same thing are equal to one another. [*Axiom* 1.

Therefore *AL* is equal to *BC*.

Wherefore *from the given point A a straight line AL has been drawn equal to the given straight line BC*. q.e.f.

## PROPOSITION 3. *PROBLEM*

From the greater of two given straight lines to cut off a part equal to the less.

Let *AB* and *C* be the two given straight lines, of which

*AB* is the greater: it is required to cut off from *AB* the greater, a part equal to C the less.

From the point *A* draw the straight line *AD* equal to *C*; [I. 2.

and from the centre *A*, at the distance *AD*, describe the circle *DEF* meeting *AB* at *E*. [*Postulate* 3.
*AE* shall be equal to *C*.

Because the point *A* is the centre of the circle *DEF*, *AE* is equal to *AD*. [*Definition* 15.

But *C* is equal to *AD*. [*Construction*.
Therefore *AE* and *C* are each of them equal to *AD*.
Therefore *AE* is equal to *C*. [*Axiom* 1.

Wherefore *from AB the greater of two given straight lines a part AE has been cut off equal to C the less.* q.e.f.

## PROPOSITION 4. *THEOREM*

If two triangles have two sides of the one equal to two sides of the other, each to each, and have also the angles contained by those sides equal to one another, they shall also have their bases or third sides equal; and the two triangles shall be equal, and their other angles shall be equal, each to each, namely those to which the equal sides are opposite.

Let *ABC*, *DEF* be two triangles which have the two sides *AB*, *AC* equal to the two sides *DE*, *DF*, each to each, namely, *AB* to *DE*, and *AC* to *DF*, and the angle *BAC* equal to the angle *EDF*: the base *BC* shall be equal to the base *EF*, and the triangle *ABC* to the triangle *DEF*, and the other angles shall be equal, each to each, to which the equal sides are opposite, namely, the angle *ABC* to the angle *DEF*, and the angle *ACB* to the angle *DFE*.

For if the triangle *ABC* be applied to the triangle *DEF*, so that the point *A* may be on the point *D*, and the straight line *AB* on the straight line *DE*, the point *B* will coincide with the point *E*, because *AB* is equal to *DE*, [*Hyp*.

And, *AB* coinciding with *DE*, *AC* will fall on *DF*, because the angle *BAC* is equal to the angle *EDF*. [*Hypothesis*.

Therefore also the point *C* will coincide with the point *F*, because *AC* is equal to *DF*. [*Hypothesis*.

But the point *B* was shewn to coincide with the point *E*, therefore the base *BC* will coincide with the base *EF*;

because, *B* coinciding with *E* and *C* with *F*, if the base *BC* does not coincide with the base *EF*, two straight lines will enclose a space; which is impossible. [*Axiom* 10.

Therefore the base *BC* coincides with the base *EF*, and is equal to it. [*Axiom* 8.

Therefore the whole triangle *ABC* coincides with the whole triangle *DEF*, and is equal to it. [*Axiom* 8.

And the other angles of the one coincide with the other angles of the other, and are equal to them, namely, the angle *ABC* to the angle *DEF*, and the angle *ACB* to the angle *DFE*.

Wherefore, *if two triangles* &c. q.e.d.

and join *FC*, *GB*.

## PROPOSITION 5. *THEOREM*

The angles at the base of an isosceles triangle are equal to one another; and if the equal sides be produced the angles on the other side of the base shall be equal to one another.

Let *ABC* be an isosceles triangle, having the side *AB* equal to the side *AC*, and let the straight lines *AB*, *AC* be produced to *D* and *E*: the angle *ABC* shall be equal to the angle *ACB*, and the angle *CBD* to the angle *BCE*.

In *BD* take any point *F*,

and from *AE* the greater cut off *AG* equal to AF the less, [I.3.

Because *AF* is equal to *AG* [*Constr.*

and *AB* to *AC*, [*Hypothesis*.

the two sides *FA*, *AC* are equal to the two sides *GA*, *AB*, each to each; and they contain the angle *FAG* common to the two triangles *AFC*, *AGB*;

therefore the base *FC* is equal to the base *GB*, and the triangle *AFC* to the triangle *AGB*, and the remaining angles of the one to the remaining angles of the other, each to each, to which the equal sides are opposite, namely the angle *ACF* to the angle *ABG*, and the angle *AFC* to the angle *AGB*. [I. 4.

And because the whole *AF* is equal to the whole *AG*, of which the parts *AB*, *AC* are equal, [*Hypothesis*.

the remainder *BF* is equal to the remainder *CG*. [*Axiom* 3.

And *FC* was shewn to be equal to *GB*;

therefore the two sides *BF*, *FC* are equal to the two sides *CG*, *GB*, each to each;

and the angle *BFC* was shewn to be equal to the angle *CGB*; therefore the triangles *BFC*, *CGB* are equal, and their other angles are equal, each to each, to which the equal sides are opposite, namely the angle *FBC* to the angle *GCB*, and the angle *BCF* to the angle *CBG*. [I. 4.

And since it has been shewn that the whole angle *ABG* is equal to the whole angle *ACF*,

and that the parts of these, the angles *CBG*, *BCF* are also equal;

therefore the remaining angle *ABC* is equal to the remaining angle *ACB*, which are the angles at the base of the triangle *ABC*. [*Axiom* 3.

And it has also been shewn that the angle *FBC* is equal to the angle *GCB*, which are the angles on the other side of the base.

Wherefore, *the angles* &c. q.e.d.

Corollary. Hence every equilateral triangle is also equiangular.

## PROPOSITION 6. *THEOREM*

If two angles of a triangle be equal to one another, the sides also which subtend, or are opposite to, the equal angles, shall be equal to one another.

Let *ABC* be a triangle, having the angle *ABC* equal to the angle *ACB*: the side *AC* shall be equal to the side *AB*.

For if *AC* be not equal to *AB*, one of them must be greater than the other.

Let *AB* be the greater, and from it cut off *DB* equal to *AC* the less, [I. 3.

and join *DC*.

Then, because in the triangles *DBC*, *ACB*, *DB* is equal to *AC*, [*Construction*.

and *BC* is common to both,

the two sides *DB*, *BC* are equal to the two sides *AC*, *CB* each to each;

and the angle *DBC* is equal to the angle *ACB*; [*Hypothesis*.

therefore the base *DC* is equal to the base *AB*, and the triangle *DBC* is equal to the triangle *ACB*, [I. 4.

the less to the greater; which is absurd. [*Axiom* 9.

Therefore *AB* is not unequal to *AC*, that is, it is equal to it.

Wherefore, *if two angles* &c. q.e.d.

Corollary. Hence every equiangular triangle is also equilateral.

## PROPOSITION 7. *THEOREM.*

On the same base, and on the same side of it, there cannot be two triangles having their sides which are terminated at one extremity of the base equal to one another, and likewise those which are terminated at the other extremity equal to one another.

If it be possible, on the same base *AB*, and on the same side of it, let there be two triangles *ACB*, *ADB*, having their sides *CA*, *DA*, which are terminated at the extremity *A* of the base, equal

to one another, and likewise their sides *CB*, *DB*, which are terminated at *B* equal to one another.

Join *CD*. In the case in which the vertex of each triangle is without the other triangle;

because *AC* is equal to *AD*,[*Hypothesis*.

the angle *ACD* is equal to the angle *ADC*.[I. 5.

But the angle *ACD* is greater than the angle *BCD*,[*Ax*. 9.

therefore the angle *ADC* is also greater than the angle *BCD*;

much more then is the angle *BDC* greater than the angle *BCD*.

Again, because *BC* is equal to *BD*,[*Hypothesis*.

the angle *BDC* is equal to the angle *BCD*.[I. 5.

But it has been shewn to be greater; which is impossible.

But if one of the vertices as *D*, be within the other triangle *ACB*, produce *AC*, *AD* to *E*, *F*.

Then because *AC* is equal to *AD*, in the triangle *ACD*, [*Hyp*.

the angles *ECD*, *FDC*, on the other side of the base *CD*, are equal to one another. [I. 5.

But the angle *ECD* is greater than the angle *BCD*, [*Axiom* 9.

therefore the angle *FDC* is also greater than the angle *BCD*;

much more then is the angle *BDC* greater than the angle *BCD*.

Again, because *BC* is equal to *BD*, [*Hypothesis*.

the angle *BDC* is equal to the angle *BCD*. [I. 5.

But it has been shewn to be greater; which is impossible.

The case in which the vertex of one triangle is on a side of the other needs no demonstration.

Wherefore, *on the same base* &c. q.e.d.

## PROPOSITION 8. *THEOREM*

If two triangles have two sides of the one equal to two sides of the other, each to each, and have likewise their

## Full text

Wikisource:The\_Elements\_of\_Euclid\_for\_the\_Use\_of\_Schools\_and\_Colleges/Book\_I.