Intriguing geometry problem regarding isogonal lines

Question

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...

Answer 1

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.

Answer 2

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.

Answer 3

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.