Question

In the adjoining 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.

Answer 1

Given: Two circles intersect each other at A and E. seg BC and seg CD are the tangents to the circles. 

To prove: □ABCD is cyclic. 

Construction: Draw AB, AE and AD. 

Proof: 

[∠EBC = ∠BAE (i) ∠EDC = ∠DAE ] (ii) [Tangent secant theorem] 

In ∆BCD,

∠DBC + ∠BDC + ∠BCD = 180° [Sum of the measures of angles of a triangle is 180°] 

∴ ∠EBC + ∠EDC + ∠BCD = 180° (iii) [B – E – D] 

∴ ∠BAE + ∠DAE + ∠BCD = 180° [From (i), (ii) and (iii)] 

∴ ∠BAD + ∠BCD = 180° [Angle addition property] 

∴ □ABCD is cyclic. [Converse of cyclic quadrilateral theorem]