Question

If a diameter of a circle bisects each of the two chords of a circle, prove that the chords are parallel.

Answer 1

Let $AB$ and $CD$ be two chords of a circle whose centre is $O$, and let $PQ$ be a diameter bisecting chords $AB$ and $CD$ at $L$ and $M$ respectively.

Since $PQ$ is a diameter. So, it passes through the centre $O$ of the circle. Now,

$L$ is mid-point of $AB$.

$OL\perp AB$ [ Line joining the centre of a circle to the mid-point of a chord is perpendicular to the chord]

$\Rightarrow \angle ALO={90}^{\circ }$

Similarly, $\angle OMD={90}^{\circ }$

Therefore, $\angle ALO=\angle OMD$

But, these are alternate interior angles.

So, $AB\parallel CD$. [Since, when the lines are parallel, then their alternate interior angles are equal.]

Hence, proved.