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Pz solve the problem with logic with correct explaination

Pz solve the problem with logic with correct explaination

Question Image
Grade:12th pass

1 Answers

Swati
37 Points
one year ago
Given : In ΔABC, P, Q and R are the mid points of sides BC, CA and AB respectively. AD ⊥ BC.
To prove : PQRD is a cyclic quardilateral.
Proof :
In ΔABC, R and Q are mid points of AB and CA respectively.
∴ RQ || BC (Mid point theorem)
Similarly, PQ || AB and PR || CA
In quadrilateral BPQR,
BP || RQ and PQ || BR (RQ || BC and PQ || AB)
∴ Quadrilateral BPQR is a parallelogram.
Similarly, quadrilateral ARPQ is a parallelogram.
∴ ∠A = ∠RPQ (Opposite sides of parallelogram are equal)
PR || AC and PC is the transversal,
∴ ∠BPR = ∠C (Corresponding angles)
∠DPQ = ∠DPR + ∠RPQ = ∠A + ∠C ...(1)
RQ || BC and BR is the transversal,
∴ ∠ARO = ∠B (Corresponding angles) ...(2)
In ΔABD, R is the mid point of AB and OR || BD.
∴ O is the mid point of AD (Converse of mid point theorem)
⇒ OA = OD
In ΔAOR and ΔDOR,
OA = OD (Proved)
∠AOR = ∠DOR (90°) {∠ROD = ∠ODP (Alternate angles) & ∠AOR = ∠ROD = 90° (linear pair)}
OR = OR (Common)
∴ ΔAOR congruence ΔDOR (SAS congruence criterion)
⇒ ∠ARO = ∠DRO (CPCT)
⇒ ∠DRO = ∠B (Using (2))
In quadrilateral PRQD,
∠DRO + ∠DPQ = ∠B + ( ∠A + ∠C) = ∠A + ∠B + ∠C (Using (1))
⇒ ∠DRO + ∠DPQ = 180° ( ∠A + ∠B + ∠C = 180°)
Hence, quadrilateral PRQD is a cyclic quadrilateral.

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