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 (see Figure). Show that: (i) Δ ABM ≅ Δ PQN (ii) Δ ABC ≅ Δ PQR

(i) Given that : BC = QR
⇒ 1/2 BC = 1/2QR ⇒ BM = QN [∵ AM and PM are medians]
In ΔABM and ΔPQN,
AB = PQ [∵ Given]
AM = PN [∵ Given]
BM = QN [∵ Proved above]
Hence, ΔABM ≅ ΔPQN [∵ SSS Congruency Rule]

(ii) In ΔABM ≅ ΔPQN, [∵ Proved above]
∠B = ∠Q [∵ CPCT]
In ΔABC and ΔPQR,
AB = PQ [∵ Given]
∠B = ∠Q [∵ Proved above]
BC = QR [∵ Given]
Hence, ΔABC ≅ ΔPQR [∵ SAS Congruency Rule]