I have a problem understanding the proof.
Given an acute angle $A$. Choose an arbitrary point $P$ from the bisector of $A$ and another point $B$ from the side of angle $A$. Draw a line $l$ going through points $P$ and $B$. Now $l$ intersects another side of the angle $A$ at point $C$. Prove that no matter how we choose $B$ and keep $P$ fixed, the expression $\frac{1}{\mid AB\mid}+\frac{1}{\mid AC\mid}$ is constant.
Proof.
Let $\angle A=2\theta$, $\angle ACB = \phi$, and $PD$ is perpendicular to $AC$. Because $AP$ bisects $\angle A$ we have ${AC\over AB}={PC\over PB}$. Thus:
${1\over AB}+{1\over AC}={1\over AB}+{PB\over AB\cdot PC}={BC\over AB}\cdot{1\over PC}={\sin 2\theta\over {PC\cdot\sin\phi}}={\sin 2\theta\over PD}={\sin 2\theta\over {PA\cdot\sin \theta}}={2\cos\theta\over PA}={\rm constant,}$
because both $\theta$ and $PA$ are given.
Why do we have that ${1\over AB}+{PB\over AB\cdot PC}={BC\over AB}\cdot{1\over PC}$ and is that proof correct?