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RHS Criteria for Congruency

In Fig. 6, AB...

Question

In Fig. 6, ABCD is a square. If ∠PQR = 90° and PB = QC = DR, prove that ∠QPR = 45°.

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Solution

$Since, ABCD is a square, then each of its sides must be equal . So, AB = BC = CD = DA Now, BC = CD Subtracting QC from both sides of the above equation, we get BC - QC = CD - QC \Rightarrow BC - QC = CD - DR (as, PB = QC = DR) \Rightarrow BQ = RC . . . . . (i) In ∆ PBQ and ∆ QCR, PB = QC (Given) ∠ PBQ = ∠ QCR = 90 ° BQ = RC [From (i)] So, by SAS congruency ∆ PBQ ≅ ∆ QCR \Rightarrow QP = QR (By CPCT) . . . . . (ii) In ∆ PQR, As, QP = QR [From (i)] \Rightarrow ∠ QRP = ∠ QPR (Angles opposite to equal sides in a ∆ are equal) Also, ∠ PQR + ∠ QRP + ∠ QPR = 180 ° (Angle sum property of ∆) \Rightarrow 90 ° + ∠ QPR + ∠ QPR = 180 ° (As, ∠ QRP = ∠ QPR) \Rightarrow 90 ° + 2 ∠ QPR = 180 ° \Rightarrow 2 ∠ QPR = 90 ° ∴ ∠ QPR = 45 °$

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