Question

Let ABCD be a square and let points P on AB and Q on DC be such that DP = AQ. Prove that BP = CQ.

Answer 1

Given: ABCD is a square and DP = AQ

To Prove: PB = CQ

Proof:

In ΔDAP and ΔADQ:

DP = AQ (Given)

∠A = ∠D = 90° (ABCD is a square)

AD = AD (Common)

⇒ ΔDAP ≅ ΔADQ (RHS congruency)

⇒ AP = DQ (Corresponding parts of congruent triangles)

Now, AB = CD (ABCD is a square)

AP + PB = DQ + QC

∴ PB = CQ