Question 6
A convex quadrilateral $ABCD$ satisfies $AB \cdot CD = BC \cdot DA$. Point $X$ lies inside $ABCD$ so that $$
\angle XAB=\angle XCD \quad \text{ and } \quad \angle XBC=\angle XDA.
$$ Prove that $\angle BXA + \angle DXC=180^\circ$.
\angle XAB=\angle XCD \quad \text{ and } \quad \angle XBC=\angle XDA.
$$ Prove that $\angle BXA + \angle DXC=180^\circ$.