Question

The given figure shows a trapezium $ABCD$ in which $AB$ is parallel to $DC$ and $AD=BC$.

Prove that : $OA=OB$ and $OC=OD$

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 $OC=OD$ [C.S.C.T.] $\left[henceproved\right]$