Join A with one of the extremities, as B, of the straight line BC (Post. 1); on AB describe the equilateral triangle ABD (i. 1); and produce the straight lines DA, DB to E, F (Post. 2). With centre B and radius BC describe (Post. 3) the circle CGH, cutting DF in &; and with centre D and radius DG describe the circle GKL, cutting DE in L. Then AL shall be equal to BC.

Because B is the centre of the circle CGH, BC is equal to BG by def"; and because D is the centre of the circle GKL, DL is equal to DG for the same reason. Now DA is equal to DB, since they are sides of the equilateral triangle DAB; from the equals DL, DG, take away the equals DA, DB: then the remainders are equal (Ax. 3), or AL is equal to BG. But BC also is equal to BG; and things that are equal to the same thing are equal to one another (Ax. 1); therefore AL is equal to BC. Hence from the given point a a straight line AL has been drawn equal to the given straight line BC. Which was to be done.

PROP. III. PROB.

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 A, that extremity of AB to which the part to be cut off is required to be adjacent, draw (i. 2) the straight line AD equal to c; and with centre A and radius AD describe (Post. 3) the circle DEF, cutting AB in E.

Then AE shall be equal to c.

A

D

B

E

Because A is the centre of the circle DEF, AE is equal to AD by def" (Def. 15). But by construction the straight line c is equal to AD; and things that are equal to the same thing are equal to one another (Ax. 3): therefore AE is equal to c. Hence from AB the greater of the two straight lines AB and C, a part AE has been cut off equal to o the less. Which was to be done.

PROP. IV. THEOREM.

If two triangles have

(1) two sides of the one respectively equal to two sides of the other;

(2) the angle included by the two sides of the one equal to the angle included by the two sides of the other:

then these triangles shall be equal in every respect; i. e. (1) the base or third side of the one shall be equal to the base or third side of the other ;

(2) the remaining angles of the one shall be respectively equal to the remaining angles of the other, those angles being equal in each to which the equal sides are opposite;

(3) the triangles shall be equal. Let ABC, DEF be two triangles, which have (1) the two sides BA, AC of the one respectively equal to the two sides ED, DF of the other, viz. BA to ED, and ac to

DF;

AA

(2) the angle BAC included by the two sides BA, AC of the one equal to the angle EDF included by the two sides ED, DF of the other.

Then these triangles shall be equal in every respect: i. e. (1) the base BC shall be equal to the base EF; (2) the remaining angles ABC, ACB shall be equal to

the remaining angles DEF, DFE respectively: viz.

ABC, DEF, to which the equal sides AC, DF are opposite, shall be equal; and ACB, DFE, to which the equal sides AB, DE are opposite, shall be equal;

(3) the triangle ABC shall be equal to the triangle

DEF.

Let the triangle ABC be applied to the triangle DEF, so that the point A may coincide with the point D, and the straight line AB may fall on the straight line DE, the triangle ABC falling on the same side of DE as the triangle DEF.

Then the point A coinciding with the point D, and the straight line AB falling on DE by const", the point B shall coincide with the point E, because AB is equal to DE by hypothesis :

Again, the straight line AB coinciding with DE, the straight line AC shall fall on DF, because the angle BAC is equal to the angle EDF by hyps; and the triangles fall by const" on the same side of DE: hence also the point c coincides with the point F, because AC is equal to DF by hyps:

But the point B was shewn to coincide with the point E; hence the point в coinciding with the point E, and the point c coinciding with the point F, the straight line BC must coincide with the straight line EF; because if it did not, it would take some other as position in the figure, and there would be two straight lines inclosing a space, which is impossible (Ax. 10). Hence the straight line BC coincides with the straight line EF; and magnitudes which coincide are equal (Ax. 8): therefore the base BC is equal to the base EF. Also the whole triangle ABC coincides with the whole triangle DEF, and the remaining angles of the one coincide with the remaining angles of the other: therefore, for the same reason as before, the angles ABC, ACB are respectively equal to the angles DEF, DFE, and the triangle ABC is equal to the triangle DEF. Hence the two triangles have been shewn to be equal to one another in every respect as was enunciated. Which was to be proved.

PROP. V. THEOR.

If a triangle be isosceles: then

(1) the angles at the base shall be equal;

(2) if the equal sides be produced, the angles on the other side of the base shall be equal.

Let the triangle ABC be isosceles, having the side AB equal to the side AC. Then

(1) the angle ABC shall be equal to the angle ACB;

(2) if the equal sides AB, AC be produced to D and E, the angles CBD, BCE on the other side of the base BC shall be equal.

D

F

B

In BD take any point F; from AE cut off (i. 3) AG equal to AF; and join CF, BG.

G

E

Because AF is equal to AG by const", and AC to AB by hyp, and the angle at A is common to the two triangles FAC, GAB; therefore these two triangles have the two sides FA, AC respectively equal to the two sides GA, AB, and the included angle FAC equal to the included angle GAB. Therefore they are equal in every respect (i. 4); and hence the base FC is equal to the base GB, and the remaining angles ACF, AGB respectively equal to the remaining angles ABG, AFC, viz. ACF to ABG, and AFC to AGB :

Again, AF is equal to AG, and AB is equal to ac; hence, taking away equals from equals, the remainder BF is equal (Ax. 3) to the remainder Co. Also, FC was shewn to be equal to CB, and the angle AFC was shewn to be equal to the angle AGB; therefore the two triangles BFC, CGB have the two sides BF, FC respectively equal to the two sides CG, GB, and the included angle BFC equal to the included angle CGB. Therefore they are equal (i. 4) in every respect; and hence the remaining angles FBC, FCB are respectively equal to the remaining angles GCB, GBC, viz. FBC to GCB, and FCB to GBC:

Now, it has been shewn that the angle ABG is equal to the angle ACF, and the angle GBC is equal to the

angle FCB: hence, taking away equals from equals, the remaining angle ABC is equal (Ax. 3) to the remaining angle ACB. And it was proved above that angle FBC is equal to angle GCB. Therefore (1) the angles ABC, ACB at the base BC are equal; (2) the angles CBD, BCE on the other side of the base BC are equal. Which was to be proved.

COR.-Every equilateral triangle shall also be equiangular.

Let ABC be an equilateral triangle. Then it shall also be equiangular.

B

A

C

Since the triangle ABC is equilateral, the side AB is equal (Def. 24) to the side AC; and therefore by the prop", the angles at the base BC, viz. ABC, ACB are equal. Again, since the triangle is equilateral, the side CA is equal to the side CB, and therefore by the prop" the angles CAB, ABC are equal. Hence each of the angles ACB, CAB is equal to the angle ABC; and things that are equal to the same thing are equal to one another (Ax. 1): therefore the angle ACB is equal to the angle CAB. Hence the three angles ABC, BCA, CAB are all equal, that is, the triangle ABC is equiangular. Which was to be proved.

PROP. VI. THEOR.

If a triangle have two of its angles equal: then the sides which subtend or are opposite to the equal angles shall be equal.

Let the triangle ABC have the angle ABC equal to the angle ACB. Then the side AC shall be equal to the side AB.

For if AB, AC be not equal, let them, if possible, be unequal, and let AB be the one which is greater than the other, AC. From BA cut off (i. 3) BD equal to AC; and join DC.

Ꭰ

A

B

C