Question

$AC$ 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

(i) It is given that $AB$ and $CD$ are the two chords of a circle

Let the point of intersection be $O$.
Join $AB,BC,CD$ and $AD$.
In triangles $AOB$ and $COD$,
$\angle AOB=\angle COD$ ...(Vertically opposite angles)

$OB=OD$ ....($O$ is the mid-point of $BD$)

$OA=OC$ ....($O$ is the mid-point of $AC$)
$△AOB\cong △COD$ ....SAS test of congruence

$\therefore AB=CD$ ....c.s.c.t.

Similarly, we can prove $△AOD\cong △BOC,$ then we get

$AD=BC$ ....c.s.c.t.

So, $\square ABCD$ is a parallelogram, since opposite sides are equal in length.

So, opposite angles are equal as well.

So, $\angle A=\angle C$

Also, for a cyclic quadrilateral opposite angles add up to ${180}^o$

So, $\angle A+\angle C={180}^o$

$\angle A+\angle A={180}^o$

$\angle A={90}^o$

So, $BD$ is the diameter. Similarly, $AC$ is also the diameter.

ii) Since $AC$ and $BD$ are diameters,

$\therefore \angle A=\angle B=\angle C=\angle D={90}^o$ ...Angle inscribed in a semi circle is a right angle

Hence, parallelogram $ABCD$ is a rectangle