Prove that $P=RA'\cap EF$, then $DP\perp EF$. In triangle $ABC$, let $DEF$ be the contact triangle, and let $(M)$ be the midpoint of the arc $(BC)$ not containing $(A)$ in $(ABC)$. Suppose ray $MD$ meets $(ABC)$ again at $R$. If $I$ is the incenter of $(ABC)$ and ray $RI$ intersects $(ABC)$ again at $A'$, then $A'$ is the antipode of $A$. If $P=RA'\cap EF$, then $DP\perp EF$.
My Progress till now: tough problem !

Lemma :Let $ABC$ be a triangle with incenter $I$, incircle $\omega$, and circumcircle $ \Omega $, and suppose that $\omega$ meets $BC, CA$, and $AB$ at $D, E,$ and $F$ . Suppose that the circle with diameter
$AI$ and $\Omega $ meet at two points $A$ and $R$. Show that $RD$ bisects angle $\angle BRC$ .
Proof : Note that the circle with diameter $AI$ will contain $E$ and $F$ .(Since $AI$ is the angle bisector and $IE=IF \implies \angle AFI=\angle AEI=90^{\circ}$ )
Note that there is a spiral symmetry $S$ centered at $R$ dilating $\Delta RFB$ to $\Delta REC$ ( considering the circle with diameter $AI$ and the circumcircle of $ABC$ ) .
So we have $\Delta KFB$ similar to $\Delta REC \implies \frac{RB}{BC}= \frac{BF}{CE}= \frac{BD}{CD}$ ( as $D,F,E$ are intoch points ) .
Hence we have ,$\frac{RB}{BC}=\frac{BD}{CD}$ and by angle bisector theorem , we get that $RD$ bisects angle $\angle BRC$ .

So, by this lemma, we get that $RD$ bisects arc $BC$ ( let's say at $M$ ).
Moreover, since $\angle AFI=\angle AEI=90^{\circ}$ , we get that $\angle ARI=90^{\circ} \implies RIA'$ are collinear , where A' is the antipode of A .
But I am stuck with point $P$.
Hope someone can give hints. Thanks in advance.
 A: 
Let $DP$ intersect circle $I$ at $K$.
Notice that $RP\times PI =FP\times PE = KP\times PD$, therefore $R,K,I,D$ are co-cyclic. Also notive $IK=ID$, therefore all the red angles are equal immediately.
Now drop a perpendicular line from $A$ to BC, we have the two pink angles being equal as $ID$ is also perpendicular to $BC$.
Since the two cyan angles are equal and the ninty degree angles are equal, we have the two green angle at vertex $A$ being equal. Therefore the top red angle is equal to the top pink angle.
Look at the red and pink angles sharing edge $ID$, we know $PD$ is parallel to $AI$. Therefore $PD$ is perpendicular to $EF$.
A: Upon looking at my old solution, I accidentally found a much nicer new solution to this problem, which I would say is "the correct way" and "the most insightful way" to this problem. So I decides to share it despite the question being so old and already answered by myself.

We denote the circumcircle of $\triangle ABC$ to be $O$ and the circumcircle of $\triangle AFE$ be $Q$.
We first notice $\angle FAE = \angle BAC$, which means arc $FE$ in $Q$ and arc $BC$ in $O$ correspond to the same angle (i.e. the two arcs have a "similar" shape in their respective circles). Secondly, $I$ is the midpoint of arc $FE$ in $Q$ because $IF=IE$ and $M$ is the midpoint of arc $BC$ in $O$. Lastly, notice that $\angle RAF = \angle RAB$, which means arc $RF$ in $Q$ corresponds to the same angle as arc $RB$ in $O$.
The above three facts directly implies quadrilaterals $RFIE$ and $RBMC$ are similar (basically they are the same shape in two different sized circles), which means ${RP\over PI}={RD\over DM}$, which means $PD$ is parallel to $IM$, which means $PD$ is perpendicular to $FE$.
