In the given figure, AB = PQ, BC = QR and median AM = median PN, then :

1. AC ≠ PR

2. BM ≠ QN

3. △ ABM ≅ △ PQN

4. △ ABC ≅ △ PQR

Given,

AM and PN are medians of triangle ABC and PQR.

∴ BM = MC and QN = NR.

BC = QR (Given)

⇒ $\dfrac{BC}{2} = \dfrac{QR}{2}$

⇒ BM = QN.

In △ ABM and △ PQN,

⇒ AB = PQ (Given)

⇒ AM = PN (Given)

⇒ BM = QN (Proved above)

∴ △ ABM ≅ △ PQN (By S.S.S. axiom)

We know that,

Corresponding parts of congruent triangles are equal.

∴ ∠AMB = ∠PNQ (By C.P.C.T.C.)

⇒ 180° - ∠AMB = 180° - ∠PNQ

⇒ ∠AMC = ∠PNR

Given,

⇒ BC = QR

⇒ $\dfrac{BC}{2} = \dfrac{QR}{2}$

⇒ MC = NR.

In △ AMC and △ PNR,

⇒ AM = PN (Given)

⇒ ∠AMC = ∠PNR (Proved above)

⇒ MC = NR (Proved above)

∴ △ AMC ≅ △ PNR (By S.A.S. axiom)

∴ AC = PR (By C.P.C.T.C.)

In △ ABC and △ PQR,

⇒ AB = PQ (Given)

⇒ BC = QR (Given)

⇒ AC = PR (Proved above)

∴ △ ABC ≅ △ PQR (By S.S.S. axiom)

Hence, Option 4 is the correct option.