Question

In the figure, $\square ABCD$ is parallelogram. Point $E$ is on the ray $AB$ such that $BE=AB$, then prove that line $ED$ bisects seg $BC$ at point F.

Answer 1

In figure,

$ABCD$ is a parallelogram.

then,

$AB||CD$

but $AB=BE$

$CD=BE$..........(i)

Now,

$\begin{array}{c}\Delta CFD\sim \Delta BFE\\ \frac{CE}{BF}=\frac{CD}{BE}\left[by\left(i\right)\right]\\ CF=BF\end{array}$

$\therefore F$ is the middle point of $BC$.