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...

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.

Projective solution:

Let perpendicular line to $$PG$$ through $$P$$ cuts $$AB$$ at $$T$$. Then $$T,A,G,D$$ are harmonic and $$T,B,G,C$$ are harmonic, so $${\vec{TB}\over \vec{BG}}: {\vec{TC}\over \vec{CG}} = -1 ={\vec{TA}\over \vec{AG}}:{\vec{TD}\over \vec{DG}}$$

and after some algebraic manipulation we get the desired equation.

Consider an inversion $$I_G$$ with center $$G$$ an arbitary radius $$R>0$$. For every point of the plane $$X$$ denote $$I_G(X)$$ as $$X'$$. Note that $$A', B', C', D'$$ will lie in order $$B', A', D', C'$$ on the line $$r$$.

Before inversion we have $$\angle GPB=\angle GPC$$ and $$\angle GPA=\angle GPD$$. Therefore, after inversion we will have $$\angle GB'P'=\angle GC'P'$$ and $$\angle GA'P'=\angle GD'P'$$. Hence, triangles $$A'PD'$$ and $$B'PC'$$ will be isosceles (with bases $$A'D'$$ and $$B'C'$$, respectively). From this we conclude that midpoints of segments $$A'D'$$ and $$B'C'$$ are coincide, so $$A'C'=B'D'$$ or $$GA'+GC'=GB'+GD'$$. Now remember that $$I_G(A)=A'$$, so $$GA\cdot GA'=R^2$$ or $$GA'=\frac{R^2}{GA}$$. Analogously, $$GB'=\frac{R^2}{GB}$$, $$GC'=\frac{R^2}{GC}$$ and $$GD'=\frac{R^2}{GD}$$. Hence, $$\frac{1}{GA}+\frac{1}{GC}=\frac{1}{GB}+\frac{1}{GD}$$, as desired.