Question

Two sides AB and BC and median AM triangle ABC are respectively equal to sides PQ and QR and median PN of APQR Show that: (i) AABM = APQN (ii) AABC = APQR ​

Answer 1

△ABC and△PQR in which AB=PQ,BC=QR and AM=PN.

Since AM and PN are median of triangles ABC and PQR respectively.

Now, BC=QR ∣ Given

⇒ 21

BC= 21

QR ∣ Median divides opposite sides in two equal parts

BM=QN... (1)

Now, in △ABM and△PQN we have

AB=PQ ∣ Given

BM=QN ∣ From (i)

and AM=PN ∣ Given

∴ By SSS criterion of congruence, we have

△ABM≅△PQN, which proves (i)

∠B=∠Q ... (2) ∣ Since, corresponding parts of the congruent triangle are equal

Now, in △ABC and△PQR we have

AB=PQ ∣ Given

∠B=∠Q ∣ From (2)

BC=QR ∣ Given

∴ by SAS criterion of congruence, we have

△ABC≅△PQR, which proves (ii)

Answer 2

Explanation:

(i) In ∆ABM and ∆PQN,

AB = PQ ...(1) | Given

AM = PN ...(2) | Given

BC = QR

⇒ 2BM = 2QN

| ∵ M and N are the mid-points of BC and QR respectively

⇒ BM = QN ...(3)

In view of (1), (2) and (3),

∆ABM ≅ ∆PQN | SSS Rule

(ii) ∵ ∆ABM ≅ ∆PQN

| Proved in (1) above

∴ ∠ABM = ∠PQN | C.P.C.T.

⇒ ∠ABC = ∠PQR ...(4)

In ∆ABC and ∆PQR,

AB = PQ | Given

BC = QR | Given

∠ABC = ∠PQR | From (4)

∴ ∆ABC ≅ ∆PQR. | SAS Rule