Suppose $AD$ is a height of $\triangle ABC$ and $H$ is the orthocenter. Is it true that $BD \cdot DC = AD \cdot DH$?

Question

Suppose $AD$ is a height of $\triangle ABC$ and $H$ is the orthocenter. Is it true that $BD \cdot DC = AD \cdot DH$?

I sense that it is false and I have tried to find a counterexample, but I cannot find it.

Could you help me to know to which subject to relate it or to demonstrate it please.

Answer 1

As mentioned by @Ananymous Triangles BDA and HDC are similar as $\angle ADB=\angle HDC=90^0$ and $\angle DAC= 90^0-(90^0-B)=B=\angle DBA$. So it follows that $$\frac{AD}{BD}=\frac{DC}{HD} \implies BD.DC=AD.HD$$