PROPOSITION 6. THEOREM.

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

A

B

Let ABC be a triangle, having the angle ABC equal to the angle ACB:

then shall the side AC be equal to the side AB.

Construction. For if AC be not equal to AB,
one of them must be greater than the other.
If possible, let AB be the greater;
and from it cut off BD equal to AC.
Join DC.

I. 3.

therefore the triangle DBC is equal in area to the triangle ACB,

1. 4.

the part equal to the whole; which is absurd. Ax. 9.
Therefore AB is not unequal to AC;
that is, AB is equal to AC.

COROLLARY.

also equilateral.

Q.E. D.

Hence if a triangle is equiangular it is

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 AC, AD, which are terminated at A, equal to one another, and likewise their sides BC, BD, which are terminated at B, equal to one another.

therefore the angle ACD is equal to the angle ADC. 1.5. But the whole angle ACD is greater than its part, the angle BCD,

therefore also the angle ADC is greater than the angle BCD; still more then is the angle BDC greater than the angle

BCD.

Again, in the triangle BCD,
because BC is equal to BD,

Нур.

therefore the angle BDC is equal to the angle BCD: but it was shewn to be greater; which is impossible.

I. 5

CASE II.

When one of the vertices, as D, is within

Then in the triangle ACD, because AC is equal to AD, Hyp. therefore the angles ECD, FDC, on the other side of the base, are equal to one another.

I. 5.

But the angle ECD is greater than its part, the angle BCD; therefore the angle FDC is also greater than the angle BCD:

still more then is the angle BDC greater than the angle

BCD.

Again, in the triangle BCD,
because BC is equal to BD,

Hyp.

therefore 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.

Therefore AC cannot be equal to AD, and at the same time, BC equal to BD.

Q.E.D.

NOTE. The sides AC, AD are called conterminous sides; similarly the sides BC, BD are conterminous.

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 bases equal, then the angle which is contained by the two sides of the one shall be equal to the angle which is contained by the two sides of the other.

Let ABC, DEF be two triangles, having the two sides BA, AC equal to the two sides ED, DF, each to each, namely BA to ED, and AC to DF, and also the base BC equal to the base EF:

then shall the angle BAC be equal to the angle EDF.

Proof. For if the triangle ABC be applied to the triangle DEF, so that the point B may be on E, and the straight line BC along EF;

then because BC is equal to EF,

Hyp.

therefore the point C must coincide with the point F.

Then, BC coinciding with EF,

it follows that BA and AC must coincide with ED and DF: for if not, they would have a different situation, as EG, GF: then, on the same base and on the same side of it there would be two triangles having their conterminous sides equal.

But this is impossible.

I. 7.

Therefore the sides BA, AC coincide with the sides ED, DF. That is, the angle BAC coincides with the angle EDF, and is therefore equal to it.

Q. E. D.

Ax. 8.

NOTE. In this Proposition the three sides of one triangle are given equal respectively to the three sides of the other; and from this it is shewn that the two triangles may be made to coincide with one another

Hence we are led to the following important Corollary.

COROLLARY. If in two triangles the three sides of the one are equal to the three sides of the other, each to each, then the triangles are equal in all respects.

The following proof of Prop. 8 is worthy of attention as it is independent of Prop. 7, which frequently presents difficulty to a beginner.

Let ABC and DEF be two triangles, which have the sides BA, AC equal respectively to the sides ED, DF, and the base BC equal to the base EF:

then shall the angle BAC be equal to the angle EDF.

For apply the triangle ABC to the triangle DEF, so that B may fall on E, and BC along EF, and so that the point A may be on the side of EF remote from D,

then C must fall on F, since BC is equal to EF. Let A'EF be the new position of the triangle ABC.

If neither DF, FA' nor DE, EA' are in one straight line,

join DA'.

CASE I. When DA' intersects EF.

Then because ED is equal to EA',

therefore the angle EDA' is equal to the angle EA'D.
Again because FD is equal to FA',

therefore the angle FDA' is equal to the angle FA'D. Hence the whole angle EDF is equal to the whole angle EA'F; that is, the angle EDF is equal to the angle BAC.

1. 5.

1. 5.

Two cases remain which may be dealt with in a similar manner: namely,

CASE II.

CASE III. line.

When DA' meets EF produced.

When one pair of sides, as DF, FA', are in one straight