Consider $BG \neq GC$ (if is equal then the problem is trivial) Denote $U$ the foot of the external bisector of $\angle BPC$ at $AD$, denote $M$ the midpoint of $UG \Rightarrow MG^2=MA.MD=MB.MC$ So $M$ is the pole of the involution that interchanges $\{A,D\},\{B,C\}$ and $G$ is fixed. So $(B,A,G,D)=(C,D,G,A) \Rightarrow \frac{BG.AD}{AG.BD}=\frac{CG.AD}{DG.AC} \Rightarrow \frac{AC}{GA.GC}=\frac{BD}{GB.GD}\Rightarrow \frac{1}{GA} + \frac{1}{GC} = \frac{1}{GB} + \frac{1}{GD} $

My solution: Lemma: Let $\triangle ABC$ be a triangle and $D$ be a point in $BC$ $\Longrightarrow$ $\frac{BD}{DC}=\frac{sin\angle BAD.sin\angle ACB}{sin\angle DAC.sin\angle BAC}$ Proof: (It's trivial by law of sines in $\triangle BAD$ and $\triangle DAC$) In the problem: Let $AB=a,BG=b,GC=c$ and $CD=d$ and let $\angle APB=\angle CPD=\theta,\angle BPG=\angle GPC=\beta,\angle PAD=\alpha$ and $\angle PDA=\gamma,\angle PGA=\omega$ $\frac{1}{GA} + \frac{1}{GC} = \frac{1}{GB} + \frac{1}{GD}$ $\Leftrightarrow$ $(a+b+c).b.(c+d)=(b+c+d).c.(a+b)$ $\Leftrightarrow$ $b.d.(a+b)=c.a.(c+d)$ Applying the lemma: $\frac{a}{b}=\frac{sin\theta.sin\omega}{sin\beta.sin\alpha}$ and $\frac{c}{d}=\frac{sin\beta.sin\gamma}{sin\omega.sin\theta}$ and $\frac{c+d}{a+b}=\frac{sin\alpha}{sin\gamma}$ $\Longrightarrow$ $\frac{a.c.(c+d)}{b.d.(a+b)}=1$ $\Longrightarrow$ $\frac{1}{GA} + \frac{1}{GC} = \frac{1}{GB} + \frac{1}{GD}$ since $\frac{1}{GA} + \frac{1}{GC} = \frac{1}{GB} + \frac{1}{GD}$ $\Leftrightarrow$ $b.d.(a+b)=c.a.(c+d)$...

This problem can also be proved by Steiners theorem ...

Steiners theorem :- If $D$ is a point on sideline $BC$ and if reflection of $AD$ in the internal angle bisector of $A$ intersects $BC$ at $E$ then :- $(BD/CD).(BE/CE)=AB^2/AC^2$.

Geometry construction See my picture consider $GP \parallel DM \parallel AN \Rightarrow (A,D,X,G)=(B,C,X,G)=-1$ (harmonic quadruples) If $AZ=AG$ and $GD=DY \Rightarrow \frac{XA}{XD}=\frac{AG}{GD}=\frac{AZ}{DY} \Rightarrow X,Y,Z $ are collinear With perspector $U$ of $(Z,Y,X,W)=(C,B,X,G)$ (harmonic quadruples) i.e. $C,U,Z$ are collinear. $\frac{1}{GU}=\frac{DB}{GB.DY}=\frac{1}{GB}+\frac{1}{GD}$ analogously $GU=\frac{1}{GA}+\frac{1}{GC}$

Attachments:

1/GA+1/GC=1/GD+1/CB if and only if 1/GB-1/GA=1/GC-1/GD if and only if AB/CD=(GB*GA)/(GC*GD). It's easy to look with the formula of areas with sines with triangles PAB and PCD, that AB/CD=(PA*PB)/(PC*PD) because they have the same heights perpendicular to AB and CD. Finally, we use two times the angle bisector theorem with triangles PAD and PBC to see that PA/PD=GA/GD and PB/PC=GB/GC, multiplying these two last we get the desired equality.

I recently found an interesting solution to this problem: First notice that $PG$ is also angle bisector of $\angle BPC$ If you play a little bit with the equality, you get that you want to prove $\frac{AC}{GA \cdot GC} = \frac{BD}{GB \cdot GD}$, that looks very similar to the formula of a segment length, if you use inversion with $G$ as the center and a convenient unit radius, then you need to prove that, under an inversion of these caracteristics, the following equality holds $A'C' = B'D'$, let's use center $G$ radius $GP=1$ now, let's see what happens: Notice that $\angle GPC = \angle GC'P$, $\angle GPB = \angle GB'P$, $\angle GPA = \angle GA'P$, and $\angle GPD = \angle GD'P$, and that $A',B',C'$ and $D'$ are still on $r$, then it follows that $PC'=PB'$ and $PA'=PD'$, simple angle chasing show that $\angle B'PA'=\angle C'PD'$ then triangles $\triangle B'PA'$ and $\triangle C'PD'$ are congruent, meaning $B'A' = C'D'$ now, it's clear that $A'B'+B'C'=A'C'$, and $D'C'+C'B'=D'B'$ so $A'C'=B'D'$ as desired

AdBondEvent wrote: Steiners theorem :- If $D$ is a point on sideline $BC$ and if reflection of $AD$ in the internal angle bisector of $A$ intersects $BC$ at $E$ then :- $(BD/CD).(BE/CE)=AB^2/AC^2$. Why man? I can´t see why this ends with this problem.

Let $PG=1$ ( if $PG \neq 1$ take a homothety of center G and radius $\dfrac{1}{PG}$ and the sum of the ratios will remain unchanged) and consider an inversion of center $G$ with radius $PG$, $PG$ bisects $\angle APD$ and $BP , CP$ are isogonal, so $\angle GPB=\angle GB'P=\angle GPC=\angle GC'P \Rightarrow PB'C'$ isosceles, analogously we arrive that $PA'D'$ is isosceles, as $\angle GPB'-\angle GPA'=\angle B'PA'=\angle GBP-\angle GAP=\angle APB=\angle CPD=\angle GCP-\angle GDP=\angle GPC'-\angle GPD'= \angle D'PC'$ we have $\triangle B'PA' \equiv \triangle D'PC' \Rightarrow A'B'=GB'-GA'=D'C'=GC'-GD'\Rightarrow \frac{1}{GA} + \frac{1}{GC} = \frac{1}{GB} + \frac{1}{GD}$