Question

In the given figure ABCD is a square. Side AB is produced to points P and Q in such a way that PA equals to AB equals to BQ. Prove that DQ equals to CP

Answer 1

In △PAD, ∠A = 90° and DA = PA = PB 

⇒ ∠ADP = ∠APD = 90° / 2 = 45°

Similarly, in △QBC, ∠B = 90° and BQ = BC = AB 

⇒∠BCQ = ∠BQC = 90° / 2 = 45°

In △PAD and △QBC , we have 

PA = QB  [given]

∠A = ∠B [each = 90°]

AD = BC [sides of a square]

⇒ △PAD ≅ △QBC [by SAS congruence rule]

⇒ PD = QC [c.p.c.t.]

Now, in △PDC and △QCD 

DC = DC [common]

PD = QC [prove above]

∠PDC = ∠QCD [each = 90° + 45° = 135°]

⇒ △PDC = △QCD [by SAS congruence rule]

⇒ PC = QD or DQ = CP

Answer 2

plz refer this pic

hope this help you!!