Question

In given figure two sides $AB,BC$ and median $AM$ of one triangle $ABC$ are respectively equal to sides $PQ$ and $QR$ and median $PN$ of $△PQR$. Show that: $△ABC\cong △PQR$

Answer 1

It is given that $AB=PQ,BC=QR$ and $AM=PN$

$\because AM$ is the median of $△ABC$

$\Rightarrow BM=CM=\frac{1}{2}BC$

$\because PN$ is the median of $△PQR$

$QN=RN=\frac{1}{2}QR$

Since, $BC=QR$

$\frac{1}{2}BC=\frac{1}{2}QR$

$BM=QN$

$\Rightarrow AB=PQ;AM=PN;BM=QN$

$\therefore △ABM\cong △PQN$ [by SSS Criteria]

$\therefore \angle B=\angle Q$ [CPCT]

In $△ABC$ and $△PQR$,

$AB=PQ$ [given]

$\angle B=\angle Q$ and $BC=QR$

$\therefore △ABC\cong △PQR$ ..... [SAS Congruence Rule]$\left[henceproved\right]$