Question

Two circles Γ and ∑ in the plane intersect at two distinct points A and B, and the centre of ∑ lies on Γ. Let points C and D be on Γ and ∑, respectively, such that C,B, and D are collinear. Let point E on ∑ be such that DE is parallel to AC. Show that AE = AB.

Answer 1

If O is the centre of ∑, then we have

But we know that ∠AEB + ∠EAB + ∠EBA = 180⁰.

Therefore

∠EBA = 180⁰ - ∠AEB -  ∠ EAB = 180⁰ - 90⁰ + 1/2∠EAB - ∠EAB = 90⁰ -1/2∠EAB:

This shows that ∠AEB = ∠EBA and hence AE = AB.