Question

If $AM$ and $CN$ are perpendiculars on the disgonal $BD$ of a parallelogram $ABCD$,Is $△AMD\cong △CNB$?Give reason.

Answer 1

In triangles $AMD$ and $CNB$

$AD=BC$(opposite sides of a parallelogram)

$\angle AMB=\angle CNB={90}^{\circ }$

$\angle ADM=\angle NBC$ since $AD\parallel BC$ and $BD$ is the transversal.

So, $△AMD\cong △CNB$ by AAS