Question

$AC=BD$

Answer 1

Draw $DM\perp AB$ and $CN\perp AB$
$\Delta DAM\equiv \Delta CBN$ by R.H.S.
$\therefore \angle DAB=\angle CBA$
$\Rightarrow \angle ADC=\angle BCD$
(Supplements of equal angles DAB and CBA).
Now by SAS, $\Delta ADB\equiv \Delta ADB$
Further, $\Delta OAD\equiv \Delta OBC$ by ASA
$\Rightarrow OA=OB$ and $OC=OD$

$\Rightarrow AC=BD$