$A, B, C$ and $D$ are on a line such that $AB=BC=CD$. Also, $P$ is a point on a circle with $BC$ as a diameter. Find $\tan\angle{APB} \cdot \tan\angle{CPD}$.
Let $O$ be the center of $(BPC)$. Let $P$' be the point of intersection of $PO$ and $(BOC)$ again. Then $C$ is the centroid of $PP'D$ since $OD$ is a median and $CD=2OD$. By symmetry,
$$\tan\angle{APB} \cdot \tan\angle{CPD}=\tan\angle{CP'D} \cdot \tan\angle{CPD}$$
This all looks promising and hopefully, it will help (or maybe it is misleading).
Thanks!
Best Answer
$\triangle$ BPC is a right angled triangle
Construct $\rightarrow$
Note that $\rightarrow$
Answer to why? :
Try to take out $\tan\angle{APB}$ and $\tan\angle{CPD}$ in terms of $PB$ and $PC$ Can you take it from here?