Similar or Congruence Triangles Theorem Proof

Statement:
If the corresponding sides of two triangles are proportional, then their corresponding angles are equal and the triangles are similar.
Two triangles ABC and DEF Such that AB/DE = BC/EF = AC/DF

To prove:
ABC similar to DEF

Construction:
Mark points P and Q on DE and DF respectively such that DP = AB and DQ = AC, join PQ.

Proof:
Since AB/DE = AC/DF and AB = DP, AC = DQ Therefore DP/DE = DQ/DF.
This implies that PQ is parallel to EF
So, < DPQ =< E and < DQP =< F (corresponding angles are equal)
Triangle DPQ similar to triangle DEF (By AA similarity) --(1)

So, DP/DE = PQ/EF Or AB/DE = PQ/EF (since DP = AB)
But AB/DE = BC/EF
BC/EF = PQ/ EF
This implies that BC = PQ

Now AB = DP, AC = DQ and BC = PQ
This implies that ABC and DPQ (SSS congruence) --2)

From 1 and 2, we get
ABC similar to DEF