Question

$ABCD$ is a parallelogram.

Answer 1

Let us draw $DN$ perpendicular to $AC$ and $BM$ perpendicular to $AC$

In $△DON$ and $△BOM$,

$\angle DNO=\angle BMO$

$\angle DON=\angle BOM$ [vertically opposite angles]

$OD=OB$ [given]

By AAS congruence rule,

$△DON\cong △BOM$

$\therefore DN=BM$

$\therefore$ Congruent triangles have equal areas

$\therefore ar\left(△DON\right)=ar\left(△BOM\right)$

If two triangles have the same base and equal areas, then they will lie between the same parallels.

$DA||CB$

In $△DOA$ and $△BOC$,

$\angle DOA=\angle BOC$ [vertically opposite angles]

$OD=OB$ [given]

$\angle ODA=\angle OBC$[alternate opposite angles]

By ASA congruence rule,

$△DOA\cong △BOC$

$\therefore DA=BC$

$\therefore$ In quadrilateral ABCD, one pair of opposite sides are equal and parallel

$\therefore ABCD$ is a parallelogram