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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.
Solution
Let O be the centre of a circle and AB, CD be the two chords.
Let PQ be the diameter bisecting chord AB and CD at L and M respectively.
L is the midpoint of AB.
So, OL ⊥ AB ⇒ ∠ ALO = 90°
Similarly,
∠CMO = 90°.
∠ ALO = ∠CMO
But these are alternate angles.
So, AB || CD ....Hence Proved.
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