line DE intersects line AB at point T.The midpoint of AB is M.Then we have TA*TB=TF*TM.You can easily prove this by Menelaus.So TP*TQ=TA*TB=TF*TM.Point M is on the circumircle of PQF.

can anyone check if this solution works? I am sort of new to moving points We will prove the following more general statement (while also switching up the labelling to be $A$-centered): Quote: The incircle of $\triangle ABC$ touches the sides $BC,CA,AB$ at $D,E,F,$ respectively. Suppose that $P$ is a point on $EF,$ and let $Q=(BCP)\cap EF.$ Prove that $M\cap (PDQ).$ Fix $\triangle ABC$ and animate $P$ on line $EF.$ Define $M'=(PDQ)\cap BC\neq D.$ Consider the following series of projective maps: Projection of $(PMD)$ to $EF$ through $A.$ ($M\to EF\cap AM$) Projection of $EF$ to $BC$ through the point at infinity along the pencil of lines perpendicular to $BC.$ ($EF\cap AM\to D$) Projection of $BC$ to $(PDQ)$ through any point. ($D\to D$) Inversion preserving $(PDQ)$ swapping $D$ and $M'.$ ($D\to M'$) The second step maps $EF\cap AM$ to $D$ through a well-known lemma. In order to show that this composition of maps is the identity map, it suffices to check that $M=M'$ for 3 cases of $P.$ Pick $P=E,F,BC\cap EF;$ the first two are easy to check by the Iran Lemma, and the last is trivial.

Let $DE\cap AB=G,\odot(P,Q,F) \cap AB=F,M$ $CF,AD,BE$are concurrent $\Rightarrow \frac{AG}{BG} =\frac{AF}{BF}$ Let$AF=a,BF=b$,so we can calculate $AG=\frac{a(a+b)}{b-a}$ $\therefore GA\cdot GB=GP\cdot GQ=GF\cdot GM$ $\Rightarrow AG(AG+a+b)=(AG+a)(AG+AM)\Rightarrow AM=\frac{a+b}{2}$ So we are done.

Attachments:

Let $K = \overline{AB} \cap \overline{DE}$. As $KA \cdot KB = KP \cdot KQ$, it suffices to show that $KA \cdot KB = KF \cdot KM$. Let $AF=x$ and $BF=z$; then because $\frac{KA}{KB} = \frac{FA}{FB} = \frac xz$, $$KA \cdot KB = \frac{xz}{(z-x)^2}(x+z)^2 = \left(\frac x{z-x} (x+z) + x\right)\left(\frac x{z-x} (x+z)+\frac{x+z}2\right) = KF \cdot KM,$$which can be checked through algebraic expansion.

McLaurin ftw. $EF \cap BC=G$ and now since $-1=(B, C; D, G)$ we have by McLaurin that $GP \cdot GQ=GA \cdot GB=GD \cdot GM$ where $M$ is the midpoint of $BC$, hence $DMPQ$ is cyclic thus we are done