Question

ABC is a triangle with ∠A  > ∠C and D is a point on BC such that ∠BAD = ∠ACB. The perpendicular bisectors of AD and DC intersect in the point E. Prove that ∠BAE = 90°.

Answer 1

∵ The perpendicular bisector of AD and DC intersect in point E 

E is the circumcentre of ∆ ADC.

Since ∠DAB = ∠ACD we have that AB is tangent to the circumcircle at A, (Alternate Segment Theroem) 

∵ radius EA ⊥ tangent AB at point of contact A, 

∴ ∠BAE = 90°.