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 ! enter image description here

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.

  • 3
    $\begingroup$ Because $\angle ARA'=\angle ARI=\angle AFI=90$ $\endgroup$ – user732848 Aug 13 '20 at 12:06
  • $\begingroup$ @Shubhangi do you know the source ? $\endgroup$ – Raheel Aug 13 '20 at 13:37

enter image description here

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 $EF$.

  • $\begingroup$ Just noticed there is no need to extend $RK$ but I have already rasterized the picture so it's hard to remove the line. I think the picture is not too confusing so I will leave it while having a comment here. $\endgroup$ – cr001 Aug 13 '20 at 14:31

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