Two sides AB and BC and median AM of one triangle ABC are respectively equal to sides PQ and QR and median PN of Δ PQR. Show that :

(i) Δ ABM ≅ Δ PQN

(ii) Δ ABC ≅ Δ PQR

Given :

AB = PQ, BC = QR = x (let) and AM = PN.

(i) Given,

AM is the median of △ ABC

∴ BM = CM = $\dfrac{1}{2}BC = \dfrac{x}{2}$ …..(1)

Also, PN is the median of △ PQR

∴ QN = RN = $\dfrac{1}{2}QR = \dfrac{x}{2}$ ……(2)

From equation (1) and (2), we get :

BM = QN ……….(3)

Now, in △ ABM and △ PQN we have,

⇒ AB = PQ (Given)

⇒ BM = QN [From equation (3)]

⇒ AM = PN (Given)

∴ △ ABM ≅ △ PQN (By S.S.S. congruence rule)

Hence, proved that △ ABM ≅ △ PQN.

(ii) Since,

△ ABM ≅ △ PQN

We know that,

Corresponding parts of the congruent triangle are equal.

∠B = ∠Q (By C.P.C.T.) ………..(4)

Now, In △ ABC and △ PQR we have

⇒ AB = PQ (Given)

⇒ ∠B = ∠Q [From equation (4)]

⇒ BC = QR (Given)

∴ △ ABC ≅ △ PQR (By S.A.S. congruence rule)

Hence, proved that △ ABC ≅ △ PQR.