Geometry Rotation and Trigonometry

geometry

$A, B, C$ 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 $\rightarrow$

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?

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