Geometry Rotation and Trigonometry 
$A, B, C$ 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!
 A: $\triangle$ BPC is a right angled triangle
Construct $\rightarrow$

1. A line parallel to $PB$ from $C$ to $PD$ at $X$ therefore$ \angle PCX$ $=90°$(why?). 
2.  A line parallel to $PC$ from $B$ to $PA$ at $Y$ therefore $\angle PBY$ $=90°$(why?)

Note that $\rightarrow$

1.  $CX=\dfrac12\cdot PB$.
2. $BY=\dfrac12\cdot PC$(why?)

Answer to why?  :

 Is the use of mid point theorem.

Try to take out $\tan\angle{APB}$ and $\tan\angle{CPD}$ in terms of $PB$ and $PC$
Can you take it from here?
A: Not an answer (please don't downvote), but I cannot imagine trying to solve this problem without drawing (or imagining) a figure such as this:

A: It is implicit from the problem that $\tan\angle{APB} \cdot \tan\angle{CPD}$ is invariant of the location of the point P. So, let P be at the top of circle. Then,
$$\tan\angle{APB} = \tan( \angle{APO} - \angle{BPO}) = \frac {\tan \angle{APO} - \tan 45 }{ 1+\tan \angle{APO}\cdot \tan45}=\frac{3-1}{1+3\cdot 1}=\frac 12$$
where O is the center of the circle and $\tan \angle APO = AO/PO=3$. Likewise, $\tan\angle{DPC} = \frac 12$. Thus, 
$$\tan\angle{APB} \cdot \tan\angle{CPD} = \frac 14$$
A: 
Let $A\equiv(0,0),B\equiv(1,0),C\equiv(2,0)$ and $D\equiv(3,0)$. Also let $\angle PBC=\theta$
$PB=\cos \theta$. Also, slope of $BP=\tan\theta$.
So, co-ordinates of $P\equiv(1+\cos^2\theta,\sin\theta \cos\theta)$
So, slope of $AP=\frac{\tan\theta}{2+\tan^2\theta}$
$tan\alpha=\frac{\tan\theta-\frac{\tan\theta}{2+\tan^2\theta}}{1+\frac{tan^2\theta}{2+\tan^2\theta}}=\frac{\tan\theta}{2}$
Slope of $CP=-\cot\theta$
$$Slope(PD)=\frac{\sin\theta\cos\theta}{\cos^2\theta-2}=-\frac{\tan\theta}{1+2\tan^2\theta}$$
similarly $\tan\beta=\frac{1}{2\tan\theta}$
