Q:

Intriguing geometry problem regarding isogonal lines

A line $r$ contains the points $A,B,C,D$ in this order. Let $P\notin r$ such that $$\angle APB=\angle CPD$$ Denote furthermore by $G$ the intersection of the angle bisector of $\angle APD$ and $r$.
Prove that $$\frac{1}{GA}+\frac{1}{GC}=\frac{1}{GB}+\frac{1}{GD}$$

My attempt so far:
Let $\Delta APD$ be a triangle, then $PB$ and $PC$ are isogonal lines. Hence $$\frac{AB}{BD}·\frac{AC}{CD}=\Bigl(\frac{AG}{GD}\Bigr)^2$$
(This fact might be proven with the Sine Law)
I don't know how to proceed now...

A:

Let $(XYZ)$ be area of triangle $XYZ$. We have $$(AGP) = (ABP)+(BGP)$$ so 
$${ab\sin \alpha\over 2}+{bg\sin \phi \over 2}={ag\sin (\alpha+\phi)\over 2}\;\;\;\;(*)$$
Law of sin for $AGP$ and $BGP$: $$ a= {AG \sin \beta \over \sin (\alpha+\phi)}\;\;\;{\rm and}\;\;\;\;b= {BG \sin \beta \over \sin (\phi)}$$
If we plug this $a$ and $b$ in to $(*)$ and rearrange we get:
$${\sin \alpha \sin \beta \over g\sin \phi \sin (\alpha+\phi)} = {1\over BG}-{1\over AG}$$
The same is true for the "right side of picture" and we are done.

