Question

Question 3 (ii)
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:
$\Delta ABC\cong \Delta PQR$

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)
$\Rightarrow \angle ABM=\angle PQN$ (By CPCT)

In $\Delta ABCand\Delta PQR$, we have
AB=PQ (Given)
$\angle B=\angle Q$ (Corresponding parts of congruent triangles $\Delta$ABM and $\Delta$PQN are equal)
BC = QR (Given)
$\therefore \Delta ABC\cong \Delta PQR$ (By SAS congruence rule)