Can anyone solve this insane hard problem?

Sorry for bumping again but I need the solution.

Let $k_1$ and $k_2$ touch $EF$ at $G$ and $L$. $R$ and $S$ are on $k_1$ and $k_2$ respectively such that $RS$ is common external tangent of $k_1, k_2$ other than $GL$. $RL$ and $GS$ intersect at $V$. $U$ is the midpoint of $I_1I_2$. $T$ is the midpoint of $RS$. $Q$ is the midpoint of $GL$. Claim: $RVUS$ is a cyclic quadrilateral. Proof: Let $(RSU)$ cut $I_1I_2$ at $V'$. We easily have $UR=US \implies I_1I_2 $ is external bisector of $\angle RUS \implies V' $ is on $GS$. $\square$ Claim: $VUAC$ is cyclic. Proof: ? Claim $A-R-L$, $G-S-C$. Proof: ? $URS \sim UAC \sim ULG$. From gliding principle $P \in QT$. Can anyone prove my claims?

Can anyone please post a diagram? I think I've misread the problem

Thanks @below, I think the problem is dependent on the diagram, because I had drawn this

Maybe an extra condition like $\overleftrightarrow{BD}$ intersects the segment $\overline{EF}$ is required Now, I just moved a point in my diagram to another position and I got the same diagram as yours @3below, Very nice solution!

Here is the diagram.

Attachments:

I hope someone can solve this problem. We spent hours on it in Michael Greenberg's live session. Then I spent hours again. This problem is really nice.

This problem is nice. Here is my solution. Let $X$ and $Y$ be the touch points of $\odot (I)$ with $AD$ and $DC$. Let $U$ and $V$ be the touch points of $\odot (I_1)$ and $\odot (I_2)$ with $EA$ and $CF$. Claim 1. $AU = CV$. Proof. First, we see that $DA + DF - AF = 2DX = 2DY = DE + DC - CE$, then we get $AF$ $+$ $AE$ $=$ $CE$ $+$ $CF$. Hence, $2AU = AE + AF - EF = CF + CE - EF = 2CV$. $\blacksquare$ Claim 2. Let $K$ and $L$ be the orthogonal projections of $I_1$ and $I_2$ on $AC$. Then $AK = CL$. Proof. Let $B$ be the orthogonal projections of $I$ on $AC$. Since $\triangle IHC \sim \triangle CLI_2$ and $\triangle IYC \sim \triangle CVI_2$, it follows that $\frac{CL}{IH} = \frac{CI_2}{CI} = \frac{CV}{IY}$. Similarly, $\frac{AK}{IH} = \frac{AU}{IX}$. Note that $AU = CV$, it implies that $AK = CL$. $\blacksquare$ Now we see that $M$ is also the midpoint of $KL$. Hence, $$ MI_1^2 - MI_2^2 = I_1K^2 - I_2L^2 = I_1A^2 - I_2C^2 = (AU^2 + r_1^2) - (CV^2 + r_2^2) = r_1^2 - r_2^2. $$It follows that $\mathcal{P}_{M/\odot (I_1)} = MI_1^2 - r_1^2 = MI_2^2 - r_2^2 = \mathcal{P}_{M/\odot (I_2)}$. 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@above Beautiful solution!