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Question

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

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Solution

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.
OLAB [ Line joining the centre of a circle to the mid-point of a chord is perpendicular to the chord]
ALO=90
Similarly, OMD=90
Therefore, ALO=OMD
But, these are alternate interior angles.
So, ABCD. [Since, when the lines are parallel, then their alternate interior angles are equal.]

Hence, proved.


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