# Triangle with Two Equal Angles is Isosceles

## Theorem

If a triangle has two angles equal to each other, the sides which subtend the equal angles will also be equal to one another.

Hence, by definition, such a triangle will be isosceles.

In the words of Euclid:

*If in a triangle two angles be equal to one another, the sides which subtend the equal angles will also be equal to one another.*

(*The Elements*: Book $\text{I}$: Proposition $6$)

## Proof 1

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.

Without loss of generality, 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 Congruence:

- $\triangle DBC = \triangle ACB$

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

Therefore, have $AB \le AC$.

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

$\blacksquare$

## Proof 2

Let $\angle ABC$ and $\angle ACB$ be the angles that are the same.

\(\text {(1)}: \quad\) | \(\ds \angle ABC\) | \(=\) | \(\ds \angle ACB\) | by hypothesis | ||||||||||

\(\text {(2)}: \quad\) | \(\ds BC\) | \(=\) | \(\ds CB\) | Equality is Reflexive | ||||||||||

\(\text {(3)}: \quad\) | \(\ds \angle ACB\) | \(=\) | \(\ds \angle ABC\) | by hypothesis | ||||||||||

\(\text {(4)}: \quad\) | \(\ds \triangle ABC\) | \(=\) | \(\ds \triangle ACB\) | Triangle Angle-Side-Angle Congruence by $(1)$, $(2)$ and $(3)$ | ||||||||||

\(\ds AB\) | \(=\) | \(\ds AC\) | from $(4)$ |

$\blacksquare$

## Sources

- 1968: M.N. Aref and William Wernick:
*Problems & Solutions in Euclidean Geometry*... (previous) ... (next): Chapter $1$: Triangles and Polygons: Theorems and Corollaries $1.13$