Given: ABCD is where and

To prove: (i) (ii)

(iii)

Proof:

(i) Here,

BC = CD …Sides of square

CQ = DR …Given

BC = BQ + CQ

∴ CQ = BC − BQ

∴ DR = BC – BQ ...(1)

Also,

CD = RC+ DR

∴ DR = CD − RC = BC − RC ...(2)

From (1) and (2), we have,

BC − BQ = BC − RC

∴ BQ = RC

(ii) Now in ∆RCQ and ∆QBP, we have,

PB = QC …Given

BQ = RC …from (i)

∠RCQ = ∠QBP …90° each

Hence by SAS congruence rule,

∆RCQ ≅ ∆QBP

∴ QR = PQ …by cpct

(iii) ∆RCQ ≅ ∆QBP and QR = PQ … from (ii)

∴ In ∆RPQ,

∠QPR = ∠QRP = (180° − 90°) = = 45°

∴ ∠QPR = 45°