Question

3. Two sides AB and BC and median AM
of one triangle ABC are respectively
equal to sides PQ and QR and median
PN of APQR (see Fig. 7.40). Show that:
(1) ABM congruent to PQN
(ii) ABC congruent to PQR​

Answer 1

Answer:

Step-by-step explanation:

Given:

AM is the median of ∆ABC & PN is the median of ∆PQR.

AB = PQ, BC = QR & AM = PN

To Show:

(i) ΔABM ≅ ΔPQN

(ii) ΔABC ≅ ΔPQR

Proof:

Since AM & PN is the median of ∆ABC

(i) 1/2 BC = BM &

1/2QR = QN

(AM and PN are median)

Now,

BC = QR. (given)

⇒ 1/2 BC = 1/2QR

(Divide both sides by 2)

⇒ BM = QN

In ΔABM and ΔPQN,

AM = PN (Given)

AB = PQ (Given)

BM = QN (Proved above)

Therefore,

ΔABM ≅ ΔPQN

(by SSS congruence rule)

∠B = ∠Q (CPCT)

(ii) In ΔABC & ΔPQR,

AB = PQ (Given)

∠B = ∠Q(proved above in part i)

BC = QR (Given)

Therefore,

ΔABC ≅ ΔPQR

( by SAS congruence rule)

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MARK ME BRAINLIEST !!

Answer 2

Explanation. in Triangle ABC, BM=1/2 BC

similarly in triangle PQR, QN=1/2 QR

As, BC=QR, therefore, BM=QN.

Now in triangle ABM and triangle PQN

AB=PQ. (given)

BM=QN (Just proved)

AM=PN (given)

therefore, triangle ABM is congruent to

triangle PQN (SSS)

so, angle ABC = angle PQR (CPCT)

Now in triangleABC and triangle PQR

AB=PQ (given)

angle ABC=angle PQR ( just proved)

BC=QR (given)

therefore triangle ABC is congruent to triangle PQR (SAS)

HENCE, PROVED......