Question

Two sides AB and BC and median AM of one triangle ABC are respectively equal to sides PQ and QR and median PN of $\Delta$PQR. Show that $\Delta$ABM $\cong$ $\Delta$PQN

Answer 1

Given:

AB = PQ,

BC = QR and

AM = PN

½ BC = BM and ½ QR = QN (Since AM and PN are medians)

Also, BC = QR

So, ½ BC = ½ QR

$\Rightarrow$ BM = QN

In $\Delta$ABM and $\Delta$PQN,

AM = PN and AB = PQ (As given in the question)

BM = QN (Already proved)

$\therefore$ $\Delta$ABM $\cong$ $\Delta$PQN by SSS congruence.