Question

In the adjacent figure $ABCD$ is a parallelogram $ABEF$ is a rectangle show that $△AFD\cong △BEC$

Answer 1

In given figure $ABCD$ is a parallelogram and $ABEF$ is a rectangle .

Then $AF=BE$ (opposite side of rectangle ).....................(1)

And $AD=BC$ (opposite side of parallelogram )...................(2)

$\angle AFE=\angle AFD=\angle BEF=\angle BEC={90}^0$ (angle of rectangle ABEF)....................................(3)

$\Delta AFD$ and $\Delta BEC$

$\angle AFD=\angle BEC={90}^0$..(From 3)

$AF=BE$ (From 1)

$AD=BC$ (From 2)

$\therefore \Delta AFD\cong \Delta BEC$