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.

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Solution

Firstly, join $O B$ .
Then, $∠ O B A = 90^{\circ}$ [Angle in a semi-circle is a right angle]
That is, $O B$ is perpendicular to $A E$ .
Now, we know that the perpendicular draw from the centre to a chord bisects the chord.
Therefore,
$A B = B E$

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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.

Firstly, join OB.Then, ∠OBA=90∘ [Angle in a semi-circle is a right angle]That is, OB is perpendicular to AE.Now, we know that the perpendicular draw from the centre to a chord bisects the chord.Therefore,AB=BE

Firstly, join $O B$ .
Then, $∠ O B A = 90^{\circ}$ [Angle in a semi-circle is a right angle]
That is, $O B$ is perpendicular to $A E$ .
Now, we know that the perpendicular draw from the centre to a chord bisects the chord.
Therefore,
$A B = B E$