Given: Triangles ΔABC and ΔDEF are such that:
∠A=∠D
To Prove: ΔABC ~ ΔDEF
Construction: Draw P and Q on DE & DF such that DP = AB and DQ = AC respectively and join PQ.
Proof:
In ΔABC and ΔDPQ
AB=DP (Construction)
∠A = ∠D (Given)
AC = DQ (Construction)
ΔABC ≅ ΔDPQ (by SAS)
∠B = ∠P (CPCT)
∠C = ∠Q (CPCT)
We can say that,
∠B = ∠P = ∠E (∠P=∠E)
∠C = ∠Q = ∠Q (∠Q=∠F)
Therefore,
In ΔABC & ΔDEF
∠B = ∠E
And, ∠C = ∠F
∴ ΔABC ~ ΔDEF
Hence Proved