$\odot I$, incircle of $\triangle ABC$ is tangent to $AB$, $BC$ and $CA$ at $F$, $D$ and $E$ respectively. $DI$ meets $AB$ at $P$. Let $FQ\perp DF$ intersecting $DE$ at $Q$. $EF$ intersects $PQ$ at $R$. If $M$ is the midpoint of $BC$, show that $RM$ is tangent to $\odot I$.
From $\angle DFQ=\angle IFA=\dfrac\pi2$, $\angle DFI=\angle QFA$. In addition, $\angle FQD=\dfrac\pi2-\angle FDQ=\dfrac12\angle A$. Since $AE=AF$, $E$, $Q$ and $F$ are all on some circle centered at $A$. So $AF=AQ$. Therefore $\triangle DFI\sim\triangle QFA$. Using this pair of similar triangles,$$\angle QAE=\angle FAQ-\angle A=\angle FID-\angle A=\pi-\angle A-\angle B=\angle C$$ so $AQ\parallel BC$. This provides a better (perhaps) definition of $Q$.
$R$ is on $EF$, the polar line of $A$ with respect to $\odot I$. So $A$ should be on that of $R$. From $\angle GFD=\dfrac\pi2$, $FQ$ and $PD$ intersect $\odot I$ at the same point, which we'll call $G$. Let $MN$ be tangent to $\odot I$. Now we only need to prove the two below:
- $RG$ is tangent to $\odot I$.
- $A$, $G$ and $N$ are collinear.
Lemma (orthogonal circles):-
Figure 1 shows two intersect circles [IEC and IEZ]. The perpendicular from C to IE is produced to cut another circle at Z. If CI is tangent to circle IEZ, then (1) IZ is a diameter of circle IEZ; and (2) ZI is tangent to circle IEC.
Produce RG to cut AEC at Z.
Then IEZ meets the description of the above lemma and therefore $\angle 4 = \angle 3$. But $\angle 3= \angle 2$ and $\angle 2 = \angle 1$ [by angle in alternate segment wrt to circle I.
Therefore, GIEZ is cyclic. This implies $\angle IGZ = 90^0$. Thus, RGZ is tangent to circle I.