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 ΔPQR (see the given figure). Show that:
$\Delta ABM\cong \Delta PQN$

Answer 1

In $\Delta$ABC, AM is the median to BC
$\therefore BM=\frac{1}{2}BC$
In $\Delta$PQR, PN is the median to QR
$\therefore QN=\frac{1}{2}QR$
However BC = QR
$\Rightarrow$ BM = QN
In $\Delta$ABM and $\Delta$PQN, we have
AB=PQ (Given)
AM=PN (Given)
BM = QN (Proved above)
$\therefore \Delta ABM\cong \Delta PQN$ (By SSS congruence rule)