Similarity of triangles, theory 4, proof On most of the internet sides I have read just 3 triangle similarity theorems, but I found out, there is also a 4: "Two triangles are similar if the lengths of two corresponding sides are proportional and their corresponding angles across the larger of these two are congruent."
I tried to find a proof for it, but I didnt find anything. I am very interested in it, I remember in high school, we studied similarity, but rarely proved anything.
 A: Proof of the SSA theorem:
Suppose you know the lengths of 2 sides of a triangle and the angle of the smaller side that is not between the 2 known sides. Without loss of generality, let the shorter side be AB and longer side be BD, with the angle extending from point A. Then, you should have a construction akin to this:

We already know the lengths of AB (shorter side) and BD (longer side). The circle with center B and radius BC is the set of all possible points satisfying the length condition of BD. Also, the known angle is $\angle BAB'$ here. Ray $\overrightarrow{AB'}$ denotes the set of all possible points satisfying the angle condition. There is only one intersection, therefore, there only exists one unique point D. This means only one unique triangle that satisfies both conditions.
Hence, if 2 triangles have the same SSA, they must be congruent. QED.
Extra: If you wonder why this fails when the angle is next to the smaller side, consider this figure:

There are now 2 intersections, meaning 2 noncongruent triangles satisfy the same conditions, so 2 triangles having the same SSA need not mean they are congruent.
