Question

Let $\sf{ABCD}$ be a convex quadrilateral such that $\sf{AB \times CD = BC \times DA}$
Let $\sf{X}$ be a point in the interior of $\sf{ABCD}$ such that $\sf{\angleXAB = \angleXCD}$ and $\sf{\angleXBC = \angleXDA}$.
Prove that $\sf{\angleBXA + \angleDXC = 180^0}$​

Answer 1

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