If P, Q and R are respectively the mid-points of sides BC, CA and AB respectively of a triangle and AD is the perpendicular from vertex A to BC, then prove that the points P, Q, R and D are cyclic.

Given that in ΔABC, P, Q and R are the mid-points of sides BC, CA and AB. Also AD ⊥ BC.

We have to prove that P, Q, R and D are concyclic.

Proof:

In ΔABC, R and Q are mid-points of AB and CA respectively.

By mid-point theorem, RQ || BC.

Also, PQ || AB and PR || CA

Consider quadrilateral BPQR,

⇒ BP ||RQ and PQ || BR

∴ BPQR quadrilateral is a parallelogram.

Similarly, ARPQ quadrilateral is a parallelogram.

We know that opposite angles of a parallelogram are equal.

⇒∠A = ∠RPQ

PR || AC and PC is the traversal,

⇒∠BPR = ∠C

⇒∠DPQ = ∠DPR + ∠RPQ = ∠A + ∠C … (1)

RQ || BC and BR is the traversal,

⇒∠ARO = ∠B … (2)

In ΔABD, R is the mid-point of AB and OR || BD.

∴ O is the mid-point of AD.

⇒ OA = OD

In ΔAOR and ΔDOR,

⇒ OA = OD

⇒∠AOR = ∠DOR = 90°

⇒ OR = OR [Common]

By SAS congruence rule,

⇒ ΔAOR ≅ Δ DRO

⇒∠ARO = ∠DRO [CPCT]

⇒∠DRO = ∠B [From (2)]

In quadrilateral PRQD,

⇒∠DRO + ∠DPQ = ∠B + (∠A + ∠C) = ∠A + ∠B + ∠C [From (1)]

Since ∠A + ∠B + ∠C = 180°,

⇒∠DRO + ∠DPQ = 180°

Hence, quadrilateral PRQD is a cyclic quadrilateral.

∴ Points P, Q, R and D are concyclic.

Hence proved.