Dear Mathlinkers, any ideas for this nice problem? Sincerely Jean-Louis

Barybash, complex seems to work pretty well, not sure about synthetic though.

Let's show that this is valid for all points on line $AB$ (using the projective plane with points at infinity, where $\ell_{\infty}$ denotes the point at infinity of line $\ell$). Let $N,Q$ and $R$ be the points of contact of $\Gamma$ with $DA,AB,BC$ respectively. Furthermore let $O$ be the center of $\Gamma$ and let $S$ be the midpoint of $NE$. Finally let $F'$ be the intersection of the tangent of $\Gamma$ at $E$ with $AD$ be $F'$. Now define two projective maps from $\Gamma$ to $AD$: the first one ($f$) maps $E\mapsto ME\mapsto MP\mapsto P\mapsto CP\mapsto F$ this is projective since it's a composition of perspectivities and a translation (line $ME$ to line $DP$). The second one ($g$) maps $E\mapsto S\mapsto F'$, and is projective since it is a composition of a homothety and a perspectivity (from circle with diameter $(ON)$ to line $AD$). Now, taking some particular case, we see $f(M)=D=g(M)$, $f(R)=\overline{BC}_{\infty}=g(R)$ and $f(Q)=A=g(Q)$. Since $f$ and $g$ are both projective maps and have coincident values for three different inputs, they are coincident maps, and in particular $F\equiv F'$ for all choices of $E$ (or equivalently $P$) and so $FE$ is always tangent to $\Gamma$ for all choices of $P$ on line $AB$.

Thank for all... And a synthetic proof? Sincerely Jean-Louis

Here's another moving points solution, which I found because I was unable to figure out a synthetic solution with the final Brokard step (oops). Let $N,Q,R$ be the respective contact points of $\Gamma$ with $DA,AB,BC$ and $\infty_{\ell}$ denote the infinity point on a line $\ell.$ Fix $ABCD$ and let $P$ move on $AB$ linearly; I claim that $PC\cap EM$ moves on the fixed line $NR.$ Observe that $PC$ clearly has degree $1+0=1$ and since $\infty_{BD}$ moves with degree $1$ on the line at infinity and $M$ is fixed, line $EM$ must have degree $1$ as well. Thus, $EM\cap CP$ has degree $1+1=2,$ and $N,R$ are fixed, so it suffices to check $2+0+0+1=3$ cases of $P:$ $P=A.$ Here, $E=Q,$ so clearly $PC,EM,NR$ meet at the center of $\Gamma.$ $P=B.$ Here, $E=R,$ so clearly $PC,EM,NR$ meet at $R.$ $P=\infty_{AD}.$ Here, $E=M,$ so clearly $PC,EM,NR$ meet at $\infty_{AD}.$ However, we know by Brokard's Theorem that $AD\cap EE,NR\cap EM, C$ are collinear, so $F=AD\cap EE$ and we're done. $\blacksquare$

Let $H$ and $G$ be the respective tangency points of the incircle with $AD$ and $BC$. Let $I = EM\cap HG$. Now observe that $\angle ADP = \angle OMI$(easy angle chase) $\implies \triangle ADP \sim \triangle OMI \implies AP=2OI$. But we also know that $CA=2CO$ so by Thales, we have $C,I,P$ are collinear. Since $F$ was on $CP$, now we know that $C,I,P,F$ are all collinear. Now by Pascal's Theorem we are done.

Dear, nice proof...just a little notation HO indeed HG... Sincerely Jean-Louis

hakN wrote: Let $H$ and $G$ be the respective tangency points of the incircle with $AD$ and $BC$. Let $I = EM\cap HG$. Now observe that $\angle ADP = \angle OMI$(easy angle chase) $\implies \triangle ADP \sim \triangle OMI \implies AP=2OI$. But we also know that $CA=2CO$ so by Thales, we have $C,I,P$ are collinear. Since $F$ was on $CP$, now we know that $C,I,P,F$ are all collinear. Now by Pascal's Theorem we are done. Which hexagon did you apply Pascal?

Let $\Gamma$ touch $AD$ and $BC$ at $J$ and $K$ respectively. Let $PC$ meet $JK$ at $L$. Since $JK$ is the $C$ midline in $\triangle CPB$, it follows that $L$ is the midpoint of segment $CP$ and consequently it lies on segment $ME$. Now by Brocard's theorem, $L$ lies on the polar line of $JE\cap KM$. By La Hire's theorem, $C$ also lies on this polar line, and hence $CL$ must be the polar of $JE\cap KM$. Consequently, the tangents at $J$ and $E$ are also concurrent with $PC$ at point $F$, so we are done.

Here's a solution using harmonic divisions and some straightforward calculations. Let $M,N$ be the point of tangency of $\Gamma$ with $(CD), (AD)$ respectively, $Z=(MF) \cap \Gamma, X=(EM)\cap (AD)$, and $a=AD$. What we want to show is: $(M,Z;N,E)=-1$. Projecting through $M$, this is equivalent to $(D,F;N,X)=-1$, or $\frac{DN}{FN}=\frac{DX}{FX}$. But clearly: $\triangle XMD \sim \triangle DPA$, so $DX=\frac{a^2}{2AP}$. Also by Thales: $\frac{AF+a}{a}=\frac{AF}{AP}$. Finally: $$\frac{DX}{FX}=\frac{DX}{DX+a+AF}=\frac{1}{1+\frac{2AF}{a}}=\frac{DN}{FN}$$,as desired.