Question

In figure, if $AB||CD$ and $E$ is the mid-point of $AC$, then show that $E$ is the midpoint of $BD$.

Answer 1

In $\Delta AEB$ and $\Delta DEC$
$AB||CD$
$\angle 1=\angle 2$ (alternate angles)
$\angle 3=\angle 4$ (vertically opposite angles)
and $AE=EC$ ($\because E$ is the mid-point of $AC$)
Similarly, $\Delta AEB\cong \Delta DEC$
(by ASA congruency property)
Hence $DE=EB$ (by c.p.c.t)
Similarly $E$ is the mid-point of $BD$.