Let us have a convex quadrilateral $ABCD$ such that $AB=BC.$ A point $K$ lies on the diagonal $BD,$ and $\angle AKB+\angle BKC=\angle A + \angle C.$ Prove that $AK \cdot CD = KC \cdot AD.$
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Tags: geometry unsolved, geometry
Let us have a convex quadrilateral $ABCD$ such that $AB=BC.$ A point $K$ lies on the diagonal $BD,$ and $\angle AKB+\angle BKC=\angle A + \angle C.$ Prove that $AK \cdot CD = KC \cdot AD.$