Question

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

Answer 1

(i) In ${\Delta }^{\prime }s$ $ABM$ and $PQN$
$AB=PQ$ (given)
$AM=PN$ (given)
and $BC=QR$ (given)
$\frac{1}{2}BC=\frac{1}{2}QR$
$BM=QN$
$\therefore \Delta ABM\cong \Delta PQN$
(by SSS congruency rule)
(ii) In $\Delta ABC$ and $\Delta PQR$
$\therefore \Delta ABM\cong \Delta PQN$ [proved in (i)]
$\angle B=\angle Q$ (by c.p.c.t)
$AB=PQ$ (given)
$BC=QR$ (given)
$\therefore \Delta ABC\cong \Delta PQR$
(by SAS congruency rule)