Question

in fig,with ∆ABC ,p,q,r are the midpoint of sides AB,AC,BC resp.

Prove that the.four triangles formed are congruent to each other.​

Answer 1

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

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.

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

Answer 2

Ello ❤

Proof:

In ∆ABC,

P, Q, R, are the midpoint of sides AB, AC, BC, resp.

By midpoint thm

PQ=1/2BC

PQ=BR=RC...... (R IS THE Midpoint OF SIDE BC) ........ (1)

lly,

PR=AQ=QC...... (2)

AND,

QR=AP=PB........ (3)

IN ∆ AP AND ∆ PBR

AP/PB=1/1

PQ/BR=1/1

AQ/PR=1/1..

From (1), (2), (3)

AP/PB=PQ/BR=AQ/PR........CORRESPONDING SIDES ARE IN PROPORTION.

∆APQ~∆PBR..... SSS TEST

LLY, ∆APQ~∆PQR

THEREFORE,

∆APQ~∆PBR~∆QCR..

HENCE PROVED