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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:

enter image description here

$\triangle \text{BCD}$ 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.
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  • $\begingroup$ The step $\angle\alpha=\angle\gamma$ uses the parallel postulate, right? $\endgroup$ – user21467 Jun 21 '15 at 14:31
  • $\begingroup$ @StevenTaschuk I'm not sure whether the parallel postulate is equivalent to corresponding-angles-fact or not. The axiom which I'm using is something like this: en.wikipedia.org/wiki/… where $\alpha _1 = \alpha$. $\endgroup$ – user103816 Jun 21 '15 at 15:43
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    $\begingroup$ I think it is. (There's also the question of how we know that $CD$ and $AB$ are parallel...) If you want to do it without the parallel postulate, you could use SSS again: you already showed that $\triangle BCD\cong\triangle BAD$ by SSS; likewise, $\triangle BAD\cong\triangle DCB$ by SSS. Therefore $\angle\beta=\angle\gamma$. (You could also skip the intermediate triangle $\triangle ABD$...) $\endgroup$ – user21467 Jun 21 '15 at 16:24
  • $\begingroup$ @StevenTaschuk Yes we can do it without parallel postulate too. I found this method on wikipedea. Is my method using parallelogram mathematically equivalent to that proof on wikipedea? $\endgroup$ – user103816 Jun 22 '15 at 7:22

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