Triangle with Two Equal Angles is Isosceles

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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

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.

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 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 $AB = AC$.

$\blacksquare$


Proof 2

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

\(\text {(1)}: \quad\) \(\ds \angle ABC\) \(=\) \(\ds \angle ACB\) Given
\(\text {(2)}: \quad\) \(\ds BC\) \(=\) \(\ds CB\) Equality is Reflexive
\(\text {(3)}: \quad\) \(\ds \angle ACB\) \(=\) \(\ds \angle ABC\) Given
\(\text {(4)}: \quad\) \(\ds \triangle ABC\) \(=\) \(\ds \triangle ACB\) Triangle Angle-Side-Angle Equality by $\left({1}\right)$, $\left({2}\right)$ and $\left({3}\right)$
\(\text {(5)}: \quad\) \(\ds AB\) \(=\) \(\ds AC\) from $\left({4}\right)$

$\blacksquare$


Sources

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