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Question
In the given figure, AC is a diameter of a circle, whose centre is O. A circle is described on AO as diameter. AE, a chord of the larger circle, intersects the smaller circle at B. Prove that : AB = BE.
Solution
Join OB,
Then ∠OBA = 90°
(Angle in a semicircle is a right angle)
i.e. OB ⊥ AE
We know the perpendicular drawn from the centre to a chord bisects the chord.
Therefore, AB = BE
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