Question

# In the given figure, two circles intersect each other at points A and E. Their common secant through E intersects the circles at points B and D. The tangents of the circles at points B and D intersect each other at point C. Prove that ▢ABCD is cyclic.

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Solution

## It is given that two circles intersect each other at points A and E. Join AE, AB and AD. The angle between a tangent of a circle and a chord drawn from the point of contact is congruent to the angle inscribed in the arc opposite to the arc intercepted by that angle. BC is the tangent to the smaller circle and BE is the chord. ∴ ∠EBC = ∠BAE .....(1) Also, CD is the tangent to the bigger circle and ED is the chord. ∴ ∠EDC = ∠DAE .....(2) Adding (1) and (2), we get ∠EBC + ∠EDC = ∠BAE + ∠DAE ⇒ ∠EBC + ∠EDC = ∠BAD .....(3) In ∆BCD, ∠DBC + ∠BDC + ∠BCD = 180º .....(4) (Angle sum property) From (3) and (4), we get ∠BAD + ∠BCD = 180º In quadrilateral ABCD, ∠BAD + ∠BCD = 180º Therefore, quadrilateral ABCD is cyclic. (If a pair of opposite angles of a quadrilateral is supplementary, the quadrilateral is cyclic) Hence proved.

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