Question

In the adjoining figure, $ABCD$ is a parallelogram.
$BM\perp ACandDN\perp AC$.
Prove that:
$\Delta BMC\cong \Delta DNA$ and,
$BM=DN$

Answer 1

$ABCD$ is a parallelogram then, we have

$AD\parallel BC$ and $AC$ is a transversal.

$\therefore$ $\angle BCA=\angle DAC$ [ Alternate interior angles ]

i.e. $\angle BCM=\angle DAN\dots \dots \left(i\right)$

In $△BMC$ and $△DNA$

$\Rightarrow$ $BC=AD$ [ Opposite sides of parallelogram are equal ]

$\Rightarrow$ $\angle BCM=\angle DAN$ [ From $eq.\left(i\right)$ ]

$\Rightarrow$ $\angle BMC=\angle DNA$ [ Both ${90}^{\circ }$, $BM\perp AC$ and $DN\perp AC$]

$\therefore$ $△BMC\cong △DNA$ [ By $AAS$ congruence rule ]

And, $BM=DN$ [ By CPCT ]