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 $JO \perp GI$

enter image description here

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.

  • $\begingroup$ 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. $\endgroup$ Mar 25, 2020 at 23:57
  • $\begingroup$ @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? $\endgroup$ Mar 26, 2020 at 3:14

1 Answer 1


We can prove $JO$ perpendicular to $GI$ in certain particular cases at least.

<span class=$JO$ 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. When <span class=$F$ 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.

When <span class=$F$ 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 $M$.

Point <span class=$F$ 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.

  • $\begingroup$ 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. $\endgroup$ Apr 4, 2020 at 2:19
  • $\begingroup$ @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. $\endgroup$ Apr 4, 2020 at 16:19

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .