IMO 2018 Problem 6

A convex quadrilateral $A B C D$ satisfies $A B\cdot C D = B C\cdot D A$ . Point $X$ lies inside $A B C D$ so that

$$\angle X A B = \angle X C D \mathrm{and} \angle X B C = \angle X D A.$$

Prove that $\angle B X A + \angle D X C = 180^{\circ}$