Question

Question 7
AC and BD are chords of a circle which bisect each other. Prove that (i) AC and BD are diameters; (ii) ABCD is a rectangle.

Answer 1

Let two chords AB and CD are intersecting each other at point O
In $\Delta AOBand\Delta COD$,
OA = OC (Given)
OB = OD (Given)
$\angle AOB=\angle COD$ (Vertically opposite angles)
$\therefore \Delta AOB\cong \Delta COD$ (SAS congruence rule)
AB = CD (By CPCT)
Similarly, it can be proved that $\Delta AOD\cong \Delta COB$
and AD = CB (By CPCT)
Since in quadrilateral ABCD, opposite sides are equal in length, ABCD is a parallelogram.
We know that opposite angles of a parallelogram are equal.
$\therefore \angle A=\angle C$
However, $\angle A+\angle C={180}^{\circ }$ (ABCD is a cyclic quadrilateral)
$\Rightarrow \angle A+\angle A={180}^{\circ }$
$\therefore 2\angle A={180}^{\circ }$
i,e $\angle A={90}^{\circ }$
As ABCD is a parallelogram and one of its interior angles is ${90}^{\circ }$, therefore, it is a rectangle.
$\angle A$ is the angle subtended by chord BD and $\angle A={90}^{\circ }$, therefore, BD should be the diameter of the circle. Similarly, AC is the diameter of the circle.