Question

In the given figure, $O$ is the middle point of both $AB$ and $CD$. Prove that $AC=BD$ and $AC||BD$.

Answer 1

In ${\Delta }^{\prime }s$ $AOC$ and $BOD$
$OA=OB$
$OC=OD$
$\angle AOC=\angle BOD$ (vertically opposite angles)
$\Delta AOC=\Delta BOD$ (by SAS congruence property)
Hence $AC=BD$ (by c.p.c.t)
Also $\angle OAC=\angle OBD$ (alt. angle)
$AC||BD$
Hence proved.