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 $ABD$?
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.