Question

Let ABC be a triangle and D be the midpoint of BC. Suppose the angle bisector of angle ADC is tangent the circumcircle of triangle ABD at D. Prove that angle A = 90 degree.

Answer 1

From the tangency we have
∠ABD = ∠BDX = ∠CDX
where X is the intersection of the angle bisector of  ∠ADC with AC
so, AB II DX
∠BAD = ∠BDX
so ABD is an isosceles triangle
BD = DA = DC
so. ∠A= 90