Triangle with Two Equal Angles is Isosceles/Proof 1

Proof

 * Euclid-I-6.png

Let $\triangle ABC$ be a triangle in which $\angle ABC = \angle ACB$.

Suppose side $AB$ is not equal to side $AC$. Then one of them will be greater.

, Suppose $AB > AC$.

We cut off from $AB$ a length $DB$ equal to $AC$.

We draw the line segment $CD$.

Since $DB = AC$, and $BC$ is common, the two sides $DB, BC$ are equal to $AC, CB$ respectively.

Also, $\angle DBC = \angle ACB$.

So by Triangle Side-Angle-Side Equality‎, $\triangle DBC = \triangle ACB$.

But $\triangle DBC$ is smaller than $\triangle ACB$, which is absurd.

Therefore, have $AB \le AC$.

A similar argument shows the converse, and hence $AB = AC$.