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 $ABD$?