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

Explanation

Let `AB \and\ CD` be two chords of a circle whose centre is `O`
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_|_AB`(Line joining the centre of a circle to the mid-point of a chord is perpendicular to the chord)
`/_ALO=90^@`
Similarly,`/_CMO=90^@`
Therefore, `/_ALO=/_CMO`
But, these are corresponding angles.
So, `AB∥CD.`