From the answers of my previous question, I got an idea to prove equality of two angles in an isosceles triangle. In that question the equality of two angles in a right-angled-isosceles triangle was shown by constructing a square consisting of two congurent triangle. The congurecy was shown by SSS. What I'hve done is that I've constructed a parallelogram from an isosceles triangle in such a way that the parallelogram consists of two isosceles triangles as shown:
$\triangle \text{BCD}$ is an isosceles triangle with $\triangle \text{BAD}$ being its replica, i.e. $\triangle \text{BCD}$ is congurent with $\triangle \text{BAD}$ by SSS. Now since $\triangle \text{BCD} \cong \triangle \text{BAD}$, we have, $\angle \delta = \angle \beta$. But $\angle \beta = \angle \alpha$ -- vertical opposite angles. And $\angle \alpha = \angle \gamma$ -- corresponding angles. This gives $$\angle \delta = \angle \gamma$$
So we have proved that the two angles of any isosceles triangle are equal.
My question is:
- Is my proof correct.
- What is it called? Who did discover it first? I want to know about its history.