Proving 4 points on a circle. Notes: I have been working on this question for a while, and I was stuck. The original question, I have already found the answer. But I wanted to try this way, and here I come. If a similar question was answered somewhere else, please link it in and close this question. Otherwise, please help me solve this
Original question
Let there be a rhombus $ABCD$. $F$ is a random point on $[AD]$.
$G, I, H$ are centers of the incircles of $\triangle ABF , \triangle DCF, \triangle BCF$. 
J is the tangent of the incircle of $\triangle BCF$ with BC. 
Prove that $JO \perp GI$ 

My attempts
What I have been trying here, I pushed the problem back to solving the following property: 
Let $K, L$ be points on $BO, CO$ such that $JK \perp BO, JL \perp CO$. Prove that $JLIG$ is inscribed in a circle ( i.e $J,L,I,G$ lies on the same circle)
Any help is appreciated.
 A: We can prove $JO$ perpendicular to $GI$ in certain particular cases at least.
$JO$ perpendicular to $GI$/">
I. Join and extend $JO$ to meet $GI$ at $E$, and join $CH$, crossing $JO$ at $L$. Now if point $F$ coincides with $A$, then since $FC$ will coincide with $AC$, and $FB$ with $AB$, the circle about $G$ is reduced to a point, and the circles about $H$ and $I$ will be tangent to each other and to diagonal $AC$ at $O$, as in the figure below.
$F$ coincides with $A$">
And since $H$ now lies on $BD$, it is clear from the equality and symmetrical placement of the circles that$$CH\parallel GI$$and hence in triangles $OLH$ and $OEI$ $$\angle LHO=\angle EIO$$And the vertical angles at $O$ are also equal.  Hence$$\triangle OLH\sim \triangle OEI$$so that$$\angle OLH=\angle OEI$$ And since $CL$ through center $H$ perpendicularly bisects chord $JO$ between the tangents, then $\angle OLH$ is right, and hence $\angle OEI$ is also right and$$JO\perp GI$$
II. At the other extreme, when $F$ coincides with $D$, then $FB$ coincides with $DB$, and $FC$ with $DC$, the circle about $I$ is reduced to a point, and the equal circles about $H$ and $G$ are tangent to each other and diagonal $BD$ at $O$, as in the next figure.
$F$ coincides with $D$[3]">
And if we join $BH$, crossing $JO$ at $M$, by the same argument as in the previous case it is clear that$$\triangle OMH\sim \triangle OEG$$and hence $JO\perp GI$.
III. Finally, take an intermediate position of $F$ where $FB=FC$. Point of tangency $J$ will now bisect $BC$, making$$JO\parallel BA$$
Extend $JE$ to $L$, and join $G$ to point of tangency at $M$.
$F$ in intermediate position">
Now since$$\triangle JOC\cong\triangle LOA$$and they are isosceles, with $LA$ tangent at $M$, then $LO$ is also a tangent.  Therefore $GI$ intersects $JL$ at tangent point $E$, and$$JO\perp GI$$
These are the two extreme cases, and just one special intermediate case. It seems a general proof will have to employ a deeper principle.
Edit: The argument in III.. above is faulty. $E$ is where $JO$ and $GI$ intersect, but when $FB=FC$ in a rhombus $JL$ is generally not tangent to the circle about $G$. Hence $E$ is generally not the point of tangency, and I have not shown $JO\perp GI$ in this particular case.
