Question

$AC$ and $BD$ are chords of a circle which bisect each other. Prove that $ABCD$ is a rectangle.

Answer 1

$AO=CO$ ....AC and BD bisect each other

$BO=DO$ ....AC and BD bisect each other

$\angle ABO=\angle COD$ ...Vertically opposite angles

$△AOB\cong △COD$ ...[SAS test]

$\Rightarrow \angle BAO$ i.e., $\angle BAC=\angle ABO$ i.e., $\angle ABD$

$\Rightarrow AC\parallel BD$
Again, $△AOD\cong △COB$ ...[SAS test]

$\Rightarrow AD\parallel CB$
$\Rightarrow ABCD$ is a cyclic parallelogram.

$\Rightarrow \angle DAC=\angle DBA$.............. (1)
[Opp.angles of a parallelogram are equal]
Also, $ACBD$ is a cyclic quadrilateral
Therefore $\angle DAC+\angle DBA={180}^{\circ }$................ (2)

From (1) and (2), we get

$\angle DAC=\angle DBA={90}^{\circ }$
Hence, $ABCD$ is a rectangle.