# 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.

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

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 $$EF$$.
• 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. – cr001 Aug 13 '20 at 14:31