Mathematics
In the given figure, ABCD is a square and P, Q, R are points on AB, BC and CD respectively such that AP = BQ = CR and ∠PQR = 90°. Prove that:
(i) PB = QC
(ii) PQ = QR
(iii) ∠QPR = 45°
Triangles
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Answer
(i) Given,
ABCD is a square.
AB = BC = CD = DA
Given,
AP = BQ = CR
From figure,
⇒ PB = AB - AP ….(1)
⇒ QC = BC - BQ
⇒ QC = AB - AP [As, BC = AB and BQ = AP] ….(2)
From eq.(1) and (2), we have :
⇒ PB = QC
Hence, proved that PB = QC.
(ii) In △PBQ and △QCR,
⇒ ∠PBQ = ∠QCR (Both equal to 90°)
⇒ BQ = CR (Given)
⇒ PB = QC (Proved above)
∴ △PBQ ≅ △QCR (By S.A.S axiom)
∴ PQ = QR (Corresponding parts of congruent triangles are equal)
Hence, proved that PQ = QR.
(iii) In △QPR,
⇒ PQ = QR
∴ △PQR is an isosceles triangle.
⇒ ∠QPR = ∠QRP = f (let) (Angles opposite to equal sides are equal)
By angle sum property of triangle,
⇒ ∠QPR + ∠QRP + ∠PQR = 180°
⇒ f + f + 90° = 180°
⇒ 2f = 180° - 90°
⇒ 2f = 90°
⇒ f =
⇒ f = 45°.
⇒ ∠QPR = ∠QRP = 45°.
Hence, proved that ∠QPR = 45°.
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