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

(N/A) In $\Delta ABC$,$AM$ is a median [Given].
$\therefore BM = \frac{1}{2} BC$ ........... $(1)$
In $\Delta PQR$,$PN$ is a median.
$\therefore QN = \frac{1}{2} QR$ ........... $(2)$
$\because BC = QR$ [Given]
$\therefore \frac{1}{2} BC = \frac{1}{2} QR$
$\Rightarrow BM = QN$ [From $(1)$ and $(2)$]
$(i)$ In $\Delta ABM$ and $\Delta PQN$,we have:
$AB = PQ$ [Given]
$AM = PN$ [Given]
$BM = QN$ [Proved above]
$\therefore \Delta ABM \cong \Delta PQN$ [$SSS$ congruence criterion]
$(ii)$ $\because \Delta ABM \cong \Delta PQN$
$\therefore$ Their corresponding parts are congruent $(CPCT)$.
$\Rightarrow \angle B = \angle Q$
Now,in $\Delta ABC$ and $\Delta PQR$,we have:
$AB = PQ$ [Given]
$\angle B = \angle Q$ [Proved above]
$BC = QR$ [Given]
$\therefore \Delta ABC \cong \Delta PQR$ [$SAS$ congruence criterion]