Question

In the adjacent figure $ABCD$ is a parallelogram and $E$ is the midpoint of the side $BC$. If $DE$ and $AB$ are produced to meet at $F$, show that $AF=2AB$

Answer 1

In given figure $ABCD$ is a parallelogram.

Here, $E$ is midpoint of $BC$. So, $BE=CE.$

$\Rightarrow$ Consider $△CDE$ and $△BFE$

$\Rightarrow$ $BE=CE$ [Given]

$\Rightarrow$ $\angle$ $CED=$ $\angle$ $BEF$ [Vertically opposite angles]

$\Rightarrow$ $\angle$ $DCE=$ $\angle$ $FBE$ [Alternate angles]

$\Rightarrow$ $△CDE\cong △BFE$ [By ASA criteria]

$\Rightarrow$ So, $CD=BF$ [CPCT] --- ( 1 )

$\Rightarrow$ But, $CD=AB$ ---- ( 2 )

$\Rightarrow$ $AB=BF$ [From ( 1 ) and ( 2 )] --- ( 3 )

$\Rightarrow$ $AF=AB+BF$

$\Rightarrow$ $AF=AB+AB$ [From ( 3 )]

$\therefore$ $AF=2AB$