# If P, Q and R are mid points of sides BC, CA and AB of a triangle ABC, and Ad is the perpendicular from A on BC, prove that P, Q , R and D are concyclic. plzz solve urgently!!!!

Given : In ΔABC, P, Q and R are the mid points of sides BC, CA and AB respectively. AD ⊥ BC.
To prove : P, Q, R and D are concyclic.
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
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  ΔDOR  (SAS congruence criterion)
⇒  ∠ARO =  ∠DRO  (CPCT)
⇒  ∠DRO = ∠B    (Using (2))
∠DRO +  ∠DPQ =  ∠B + ( ∠A +  ∠C) =  ∠A +  ∠B +  ∠C  (Using (1))
⇒  ∠DRO +  ∠DPQ = 180°  ( ∠A +  ∠B +  ∠C = 180°)
Thus, the points P, Q, R and D are concyclic.
• 50

Given : In ΔABC, P, Q and R are the mid points of sides BC, CA and AB respectively. AD ⊥ BC.

To prove : P, Q, R and D are concyclic.

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

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  ΔDOR  (SAS congruence criterion)

⇒  ∠ARO =  ∠DRO  (CPCT)

⇒  ∠DRO = ∠B    (Using (2))

∠DRO +  ∠DPQ =  ∠B + ( ∠A +  ∠C) =  ∠A +  ∠B +  ∠C  (Using (1))

⇒  ∠DRO +  ∠DPQ = 180°  ( ∠A +  ∠B +  ∠C = 180°)

Thus, the points P, Q, R and D are concyclic.

• 19

this means degree (o)

• -11

Given : In ΔABC, P, Q and R are the mid points of sides BC, CA and AB respectively. AD ⊥ BC.

To prove : P, Q, R and D are concyclic.

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

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  ΔDOR (SAS congruence criterion)

⇒ ∠ARO = ∠DRO (CPCT)

⇒ ∠DRO = ∠B (Using (2))

∠DRO + ∠DPQ = ∠B + ( ∠A + ∠C) = ∠A + ∠B + ∠C (Using (1))

⇒ ∠DRO + ∠DPQ = 180° ( ∠A + ∠B + ∠C = 180°)

Thus, the points P, Q, R and D are concyclic.

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s
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• 8
Easyly solve

• -11
Hope this will be useful

• 7

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?

?

Given : In ?ABC, P, Q and R are the mid points of sides BC, CA and AB respectively. AD ? BC.

To prove : P, Q, R and D are concyclic.

Proof :

In ?ABC, R and Q are mid points of AB and CA respectively.?

? RQ || BC ?(Mid point theorem)

Similarly, PQ || AB and PR || CA

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 ?ABD, 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 ?AOR and ?DOR,

OA = OD ?(Proved)

?AOR = ?DOR ?(90?) ?{?ROD = ?ODP ?(Alternate angles) & ??AOR = ?ROD = 90? ?(linear pair)}

OR = OR ?(Common)

? ?AOR ??DOR ?(SAS congruence criterion)

? ??ARO = ??DRO ?(CPCT)

? ??DRO = ?B ? ?(Using (2))

??DRO + ??DPQ = ??B + ( ?A + ??C) = ??A + ??B + ??C ?(Using (1))

? ??DRO + ??DPQ = 180? ?( ?A + ??B + ??C = 180?)

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