Question

In the given figure, ABCD is a square and ∠PQR = 90°. If PB = QC = DR, prove that
(i) QB = RC,
(ii) PQ = QR,
(iii) QPR = 45°.

Answer 1

Given: ABCD is a square and ∠PQR = 90°.
Also, PB = QC = DR

(i) We have:
BC = CD (Sides of square)
CQ = DR (Given)
BC = BQ + CQ
⇒ CQ = BC − BQ
∴ DR = BC − BQ ...(i)

Also, CD = RC+ DR
∴ DR = CD − RC = BC − RC ...(ii)
From (i) and (ii), we have:
BC − BQ = ​BC − RC
∴ BQ = RC

(ii) In ∆RCQ and ∆QBP, we have:
PB = QC (Given)
BQ = RC (Proven above)
∠RCQ = ∠QBP (90^o each)
i.e., ∆RCQ ≅ ∆QBP (SAS congruence rule)
∴ QR = PQ (By CPCT)

(iii) ∆RCQ ≅ ∆QBP and QR = PQ (Proven above)
∴ In ∆RPQ, ∠QPR = ∠QRP = $\frac{1}{2}\left(180°-90°\right)=\frac{90°}{2}={45}^{°}$