In the quadrilateral $ABCD$ it is known that $\angle ABD= \angle DBC$ and $AD= CD$. Let $DH$ be the altitude of $\vartriangle ABD$. Prove that $| BC - BH | = HA$. (Hilko Danilo)
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Tags: geometry, equal segments, equal angles
In the quadrilateral $ABCD$ it is known that $\angle ABD= \angle DBC$ and $AD= CD$. Let $DH$ be the altitude of $\vartriangle ABD$. Prove that $| BC - BH | = HA$. (Hilko Danilo)