Question

If a diameter of a circle bisect each of the two chords of a circle than prove that the chords are parallel.

Answer 1

Let AB and CD be the two chords of the circle, PQ be the diameter and O be the centre of the circle.
Let PQ divide the two chords AB and CD, respectively at M and N.
Therefore, we can say that OM divides AB.

We know that the perpendicular from the centre on a chord divides the chord.

So, by the converse of this, OM $\perp$ AB.
That is, $\angle \mathrm{OMB}=90°$.

Similarly, $\angle \mathrm{OND}=90°$.

$\therefore \angle \mathrm{OMB}+\angle \mathrm{OND}=90°+90°=180°$

The sum of angles on the same sides of the transversal line are supplementary.

Therefore, chord AB and CD are parallel.