Question 76
Point $D$ inside an acute triangle $ABC$ satisfies
$$
\angle ADC = \angle BDA = 180^{\circ} - \angle CAB.
$$
Prove that the point symmetric to point $A$ w.r.t. point $D$ lies on the circumcircle of the triangle $ABC$.
$$
\angle ADC = \angle BDA = 180^{\circ} - \angle CAB.
$$
Prove that the point symmetric to point $A$ w.r.t. point $D$ lies on the circumcircle of the triangle $ABC$.