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Chord of the Bigger Circle Is Bisected at the Point of Contact with the Smaller Circle

In the given ...

Question

In the given figure, circles with centres C and D touch internally at point E. D lies on the inner circle. Chord EB of the outer circle intersects inner circle at point A. Prove that, seg EA ≅ seg AB.

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Solution

Circles with centres C and D touch internally at point E.

Join ED.

By theorem of touching circles, points E, C and D are collinear.

Since D lies on the inner circle with centre C, therefore, ED is the diameter of the inner circle.

∴ ∠EAD = 90º (Angle inscribed in a semi-circle is a right angle)

EB is the chord of the outer circle with centre D.

∴ Point A is the mid-point of chord EB. (Perpendicular drawn from the centre of a circle on its chord bisects the chord)

⇒ seg EA ≅ seg AB

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