**Side-Angle-Side Similarity
Theorem**

If A, B, C are distinct points, and A', B', C' are distinct points, and k > 0 is a real number, with kAB = A'B', kAC = A'C', and a(B,A,C) = a(B',A',C'), then the triangles ABC and A'B'C are similar, with similarity ratio k.To prove the SAS Similarity Theorem we'll use the theorem from Euclid's Elements we discussed in class (Book VI, Proposition 2):

**Proportional Sides Theorem**

Let A, B, C be noncollinear points, let D be between A and B, and let E be between A and C. The points D and E divide the sides AB and AC proportionally (BD/AD = CE/AE) if and only if the line DE is parallel to the line BC.

To apply the Proportional Sides Theorem, we use a little algebra fact.

**Lemma**

With the same hypotheses as in the Proportional Sides Theorem, BD/AD = CE/AE if and only if AB/AD = AC/AE.

To prove this lemma, let x = AD, y = BD, u = AE, v = CE. Then the statement AB/AD = AC/AC translates to

(x+y)/x = (u+v)/u1 + y/x = 1 + v/u

y/x = v/u

This last statement is equivalent to BD/AD = CE/AE. This proves the lemma.

Now here is the proof of the SAS Similarity Theorem.

We are given AB/A'B' = AC/A'C' and a(B,A,C) = a(B',A',C'). Let D be the point between A and B such that AD = A'B', and let E be the point between A and C such that and AE = A'C'. Then triangle ADE is congruent to triangle A'B'C' by the SAS congruence axiom. So we just have to prove that triangles ABC and ADE are similar to deduce that triangles ABC and A'B'C' are similar.

Now AB/AD = AC/AE. Thus BD/AD = CE/AE by the algebra lemma. Therefore DE is parallel to BC by the Proportional Sides Theorem. It follows from the corresponding angles theorem that a(A,D,E) = a(A,B,C) and a(A,E,D) = a(A,C,B).

It remains to prove that BC/DE = AB/AD. Let L be the line through D parallel to CE, and let F be the intersection of L with BC. Thus DFCE is a parallelogram. It follows that DE = FC. Therefore BC/DE = BC/FC. But BC/FC = AB/AD by the Proportional Sides Theorem. This completes the proof.