other; their third sides will be equal; and their surfaces will be equal; and their other angles to which the equal sides are opposite will be equal, each to each.*

Let ABC, DEF be two triangles, which have the side AB equal to DE, and AC to DF; and the angle A equal to D; then the base BC will be equal to EF, and the triangle ABC to DEF; and the other angles, to which the equal sides are opposite, will be equal, each to each, namely, the angle B to E, and the angle C to F.

If the triangle ABC be applied to the triangle DEF, so that the point A may be on D, and the straight line AB on DE, the point B will coincide with E, because AB is equal to DE; and because AB coincides with DE, and the angle A is equal to D, AC will coincide with DF; wherefore also the point C will coincide with F, because AC is equal to DF. But the point B coincides with E; wherefore the base BC will coincide with EF, and will be equal to it. Therefore the whole triangle ABC will coincide with the whole triangle DEF, and be equal to it; and the remaining angles of one triangle will coincide with the remaining angles of the other, and be equal to them, namely, the angle B to E, and the angle C to F.

Therefore, if two triangles have two sides of one triangle equal to two sides of the other, each to each, and have also the angles contained by those sides equal to each other; their bases will be equal, and their surfaces will be equal, and their other angles to which the equal sides are opposite will be equal, each to each. Which was to be demonstrated.

* Shorter enunciation, thus:-If two sides of one triangle be equal to two sides of another, each to each, and if the angles contained by those sides be also equal, the triangles will be equal in all respects.

PROPOSITION A. THEOREM.

This is Prop. 26.

If two triangles have two angles of one triangle equal to two angles of the other, each to each, and have also the sides between those angles equal to each other; then will the other two sides be equal, each to each, and the third angle of one triangle to the third angle of the other.

Let ABC, DEF be two triangles which have the angles B, C equal to the angles E, F, each to each, and the side BC equal to EF; then the other sides will be equal, each to each, namely, AB to DE, and AC to DF; and the third angle A to the third angle D.

CE

Let the triangle ABC be applied to DEF, so that the side BC may coincide with EF; then the side AB will lie on DE, because the angle B is equal to E, and the side AC will lie on DF, because the angle C is equal to F. Therefore the point A where BA, CA meet, will fall on the angular point D. Consequently the side AB coincides with DE, AC with DF, the angle A with D, and the triangle ABC with DEF. Therefore, if two triangles &c. Q. E.D.

ED.

Cor. 1. if two angles of a triangle be equal to each other, the sides opposite to them are equal."

For suppose the two angles at the base of each triangle to become equal to each other, then the four angles at the bases of both triangles will be equal to one another; therefore the sides opposite to them will be equal to one another by the proposition,

E

ED.

Cor. 2. If two triangles be mutually equiangular, and have two corresponding sides equal to each other, the other corresponding sides will be equal, and the triangles will be equal in all respects. ED.

PROPOSITION V. THEOREM.

IF two sides of a triangle be equal, the angles opposite to them are equal.

Let ABC be a triangle, of which the side AB is equal to AC; the angle ABC is equal to ACB.

Produce AB, AC to D and E. In BD take any point F, and from AE, the greater, cut off AG equal to AF, the less (3. 1.), and join FC, GB.

The side AF is equal to AG, and AB to AC, and the angle A is common to the two triangles AFC, AGB; therefore the base FC is equal to GB (4. 1.), and the angle ACF to ABG, and the angle AFC to AGB.

Because AF is equal to AG, and the part AB to AC, the remainder BF is equal to the remainder CG (3 Ax.)

In the triangles BFC, CGB, the side BF is equal to CG, and FC to GB, and the angle BFC to

D

A

B

C

F

G

E

CGB; therefore the angle BCF is equal to CBG (4. 1.).

Now, since it has been proved that the angle ABG is equal to ACF, and the part CBG to the part BCF, the remaining angle ABC is equal to the remaining angle ACB (3 Ax.). Therefore, the angles at the base &c. Q. E. D.* :

Otherwise. If the side AB be equal to AC, the angle ABC is equal to ACB. For no reason can be assigned that the angle ABC should be either greater or less than ACB; therefore the angles ABC, ACB are equal. Q. E. D.

ED.

Otherwise. This proposition is merely a corollary to the fourth.

*Quod erat demonstrandum, which was to be demonstrated.

For, let the sides AB, AC be equal to each other, and also the sides DE, DF be equal; then the four sides AB, AC, DE, DF, are equal to one another; therefore the four angles subtended by those four sides are equal to one another (4, 1.). Therefore the two angles at the base of each triangle are equal to each other. Q. E. D.

ED.

Cor. Hence every equilateral triangle is also equiangular,

Let ABC be an equilateral triangle; then because the two sides AB, AC are equal, the angles B and C at the base are equal. Again, because the sides BA, BC are equal, the angles A and C at the base are equal. Hence each of the angles B and A is equal to C; therefore they are equal to each other (1 Ax.). B Wherefore the three angles of the triangle ABC are equal, that is, the triangle ABC is equiangular,

ED.

A

PROPOSITION VI. THEOREM.*

If two angles of a triangle be equal, the sides opposite to them are equal. See Note:

Let ABC be a triangle having the angle B equal to C; the side AB is also equal to AC.

D

A

For if AB be not equal to AC, one of them is greater than the other. Let AB be the greater, and from it cut off DB equal to AC, the less (3. 1.), and join DC. In the triangles DBC, ACB, DB is equal to AC, and BC common to both, and the angle B is equal to ACB (by the supposition); therefore the triangle DBC is equal to ACB (4.1.), the less to the greater, which is absurd. Therefore the side AB is not greater than AC. In the same manner it may be proved that AB is not less than AC. Consequently AB is equal to AC. Therefore, if two angles &c, Q. E. D.

B

C

Otherwise. If the angle B be equal to C, the side AB is equal to AC. For no reason can be assigned that the side AB should be either greater or less than AC; therefore the sides AB AC are equal. Q. E.D.

ED.

Cor. Hence every equiangular triangle is also equilateral.

Let ABC be an equiangular triangle (figure to Cor. Prop. V); then because the angle B is equal to C, the side AC is equal to AB; and because the angle A is equal to C, the side BC is equal to AB. Hence each of the sides AC and BC is equal to AB; therefore they are equal to each other (1 Ax.). Therefore the three sides of the triangle ABC are equal, that is the triangle ABC is equilateral. ED.

*This is the converse of Proposition V, and may be inferred from it without demonstration.