# 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.

• Do you mean prove $K, L, I, G$ lie on the same circle? They appear to, but how will that help? Wouldn't it be better to prove $J, L, E, G$ are on the same circle? ($E$ is the point where $JO$ meets $GI$.) If you could show that $\angle OJL=\angle OGE$ that would do it. Mar 25, 2020 at 23:57
• @EdwardPorcella well, I have pushed the problem to just prove $K,L,I,G$ lie on the same circle. I think $J,L,E,G$ works too, but I haven't know how. So why don't you just give me a solution to my question and I will give you how I finish it? Mar 26, 2020 at 3:14

## 1 Answer

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.

• The original proof is what I wouldn't say, much easier, but what I would say, much shorter. And about your answer, totally fine. But it is not going my way, which is proving the 4 points $K,L,I,G$ lies on the same circle. I shall give here the original answer if possible so people don't follow that route anymore and help with my task. Apr 4, 2020 at 2:19
• @Minh Trung Dang--I have not yet been able to prove $K, L, I , G$ lie on the same circle. The above was the best I could do at showing $JO$ perpendicular to $GI$. It does not follow the path you wanted to take, and I'm afraid the proof for the third case does not work anyway: $E$ is the intersection of $JL$ and $GI$, and $JL$ is tangent to the circle around $G$, but I have not shown that $E$ is the point of tangency. Does the answer you mention use more than Euclidean geometry? I'll give some more thought to proving $K, L, I, G$ are concyclic. Good problem. Apr 4, 2020 at 16:19