Flag 6 grade maths> Pz solve the problem with logic with corr...
question mark

Pz solve the problem with logic with correct explaination

amit arya , 8 Years ago
Grade 12th pass
anser 1 Answers
Swati

Last Activity: 4 Years 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.

Provide a better Answer & Earn Cool Goodies

Enter text here...
star
LIVE ONLINE CLASSES

Prepraring for the competition made easy just by live online class.

tv

Full Live Access

material

Study Material

removal

Live Doubts Solving

assignment

Daily Class Assignments


Ask a Doubt

Get your questions answered by the expert for free

Enter text here...