Question

In the figure, ABCD is a parallelogram is produced to E such that $AB=BE$. AD produced to F such that $AD=DF$. Show that $△FCD\cong △CBE$.

Answer 1

In given figure

$ABCD$ is a parallelogram

$\Rightarrow \angle CBA=\angle CDA$ ....... (opposite angle of parallelogram)

$\therefore {180}^o-\angle CBA={180}^o-\angle CDA$

$\therefore \angle CBE=\angle FDC$ .......... (1)

Now, $AB=DC$ ..... (opposite sides of parallelogram)

$AB=BE$ ...... (given)

$\therefore BE=DC$................(2)

Also, $BC=AD$ ...... (opposite sides of parallelogram)

and $BC=DF$ .......... (given)

$\therefore BC=DF$....................(3)

From $\left(1\right),\left(2\right)$ and $\left(3\right)$

$△CBE\cong △FCD$ ......... [By SAS congruence rule]