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I was doing the following problem

An isoceles triangle is a triangle in which two sides are equal. Prove that the angles opposite to the equal sides are equal.

I drew this diagram (sorry for the large picture):

Name the triangle $ABC$ such that $AB=BC$. The angle bisector of $\angle B$ intersects line $AC$ at the point $D$. Now we have the two triangles $ABD$ and $CBD$. They both share side $BD$, $m\angle ABD = m\angle CBD$, and by hypothesis $AB=BC$. so the two triangles are congruent by $SAS$.

So what I know so far is that there exists a way to place triangle $ABD$ onto triangle $BDC$ so that they overlap perfectly. From the picture it is clear that this could be achieved by reflecting triangle $BDC$ over the line $BD$, which would imply the desired conclusion. But is this sort of "from the picture" argument really valid/rigorous? What if the actual way to make the triangles overlap is to put side $BD$ of triangle $BDC$ onto side $AB$ of triangle $ABD$?

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    $\begingroup$ But is this sort of "from the picture" argument really valid/rigorous? No, it is not unless you back it up with some more formal proof. In this case, remember that the triangles equality previously proved implies that $AD=DC$ and $\angle BDA=\angle BDC = 90^\circ\,$, then using those equalities you could justify the "flip over" argument (though at that point you already know that $\angle A = \angle C$ by the same triangle equality). $\endgroup$ – dxiv Aug 15 '17 at 1:07
  • $\begingroup$ FWIW the more direct proof is that $\triangle BAC = \triangle BCA$ by $SSS\,$ (or $SAS\,$), therefore the corresponding angles are equal $\angle BAC = \angle BCA$. $\endgroup$ – dxiv Aug 15 '17 at 1:20
  • $\begingroup$ Let me ask you, how is it clear from the picture that you can do that? It can only be clear if you are somehow certain that AD = DC and AB=AC , BD = BD $\angle A = \angle C; \angle BAD =\angle CAD; \angle BDA = \angle CDA$ etc. Can you be certain of that? If you can, then yes. If not then, no. But if yes, you must be able to say why. $\endgroup$ – fleablood Aug 15 '17 at 1:34
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Yes, you are right that "there exists a way to place triangle $ABD$ onto triangle $BDC$ so that they overlap perfectly", and you might argue: "well there are more than one way to place, how do I know which way they overlap perfectly" - and that's exactly your question.

However, do not forget that you get the congruence by $SAS$, which means that you know which two sides are equal respectively, and you also know which angle (i.e. the angle of those two equal sides) are the same for the two triangles.

Then, it is obvious which way you should place the triangles.

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Even if your proof is the preferred one in high-school textbooks, I think a simpler proof is worth mentioning.

Compare triangles $ABC$ and $CBA$: they are congruent by $SAS$ and thus $\angle A\cong\angle C$.

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