Assume WLOG $AB < AC$. $\measuredangle RBS = \measuredangle RSB= \measuredangle IRS$. On the other hand $\measuredangle PRA= \measuredangle BRS$. And $\measuredangle AIP+ \measuredangle RIP = \measuredangle AIR= \measuredangle PAI+ \measuredangle SBC$, $( \star )$. Law of sines in $\triangle PAI \Longrightarrow \frac{PA}{PI} = \frac{ \sin \angle AIP}{ \sin \angle PAI}$ Law of sines in $\triangle PIR \Longrightarrow \frac{PI}{PR} = \frac{ \sin \angle PRI}{ \sin \angle RIP}$ Law of sines in $\triangle PRA \Longrightarrow \frac{PR}{PA} = \frac{ \sin \angle PAR}{ \sin \angle PRA}$ Law of sines in $\triangle RBS \Longrightarrow \frac{BS}{BR} = \frac{ \sin \angle BRS}{ \sin \angle RSB}$ Law of sines in $\triangle BSC \Longrightarrow \frac{SC}{BS} = \frac{ \sin \angle SBC}{ \sin \angle ACB}$ Multiplying all the equations we get: $\frac{ \sin \angle PAI}{ \sin \angle SBC} = \frac{ \sin \angle AIP}{ \sin \angle RIP}$, $( \star \star )$. By $( \star )$, $( \star \star )$ and Two Equal Angles Lemma it follows $\measuredangle PAI= \measuredangle AIP$. We conclude $PI=PA$.

Here's another solution: Let $E = RI \cap AS$, $F = SI \cap AR$, $O$ and $J$ the circumcenter and $A$-excenter of triangle $\triangle ARS$, respectively. Let $\phi$ a inversion centered at $A$ and radius $\sqrt{AR \times AS}$ followed by a reflection on angle bissector of $\angle RAS$ Since $AB=AR+RS$ and $AE=\frac{AR \times AS}{AR+RS} \Longrightarrow \phi(B)=E$. Analogously, $\phi(C)=F$. Also we have $BC \parallel EF, \phi(I)=J$ and $\phi(AP)$ is the line $l$ through $A$ parallel to $BC$. Let $K \neq A$ the intersection of $l$ and the circumcircle of triangle $\triangle ARS$. Then, $\phi(P)=K$ Then, $PA=PI \Longleftrightarrow \angle PAI=\angle PIA \Longleftrightarrow \angle KAJ=\angle AKJ \Longleftrightarrow JA=JK \Longleftrightarrow JO \perp AK \Longleftrightarrow OJ \perp EF$. This is well-known, see here: https://www.artofproblemsolving.com/community/c5h1230491p6213572 We are Done!

Lemma: In a $\triangle ABC$ with orthocenter $H$ and 9-point center $N,$ let $D,E,F$ be the feet of the altitudes on $BC,CA,AB.$ The perpendicular bisector of $\overline{DH}$ cuts $EF$ at $L.$ Then $LH \perp AN.$ Proof: Let the perpendicular from $H$ to $AN$ cuts $EF$ at $L'$ and $\odot(HBC)$ again at $X.$ As the inversion with center $H$ and power $HA \cdot HD=HB \cdot HE=HC \cdot HF$ takes $\odot(HBC)$ into $EF,$ it follows that $AXDL'$ is cyclic $\Longrightarrow$ $\triangle HAX \sim \triangle HL'D.$ But since $AN$ passes through the center of $\odot(HBC)$ (reflection of $A$ on $N$), then $\triangle HAX$ is A-isoscles $\Longrightarrow$ $\triangle HL'D$ is L'-isosceles $\Longrightarrow$ $L \equiv L'$ $\Longrightarrow$ $LH \perp AN. \ \blacksquare$ Back to the proposed problem. Redefine $P$ as the intersection of $RS$ with the perpendicular bisector of $\overline{AI},$ thus we shall show that $PA$ is tangent to $\odot(ABC).$ Let $U$ and $J$ be the circumcenter and A-excenter of $\triangle ARS.$ Using the previous lemma on the excentral triangle of $\triangle ARS,$ it follows that $ PI \perp JU.$ But by extraversion of a well-known property discussed at https://artofproblemsolving.com/community/c6h321293 (see post #5), we have $JU \perp BC,$ $\Longrightarrow$ $IP \parallel BC$ $\Longrightarrow$ $\angle PAI=\angle PAB+\tfrac{1}{2}\angle BAC=\tfrac{1}{2}\angle BAC+\angle ACB$ $\Longrightarrow$ $\angle ACB=\angle PAB$ $\Longrightarrow$ $PA$ is tangent to $\odot(ABC).$

uh..... what problem is this?

Here is a simple solution. Let $\triangle DEF$ be the cevian triangle of $I$ w.r.t $\triangle ARS$, $EF$ meets $RS$ at $T$. Since $SF$ is the bisector of $\angle ASR$ so $SF$ $\parallel $ $RC$, thus $\tfrac{AF}{AR}$ $=$ $\tfrac{AS}{AC}$ imply $AF\cdot AC$ $=$ $AR\cdot AS$. Similarly, $AE\cdot AB$ $=$ $AR\cdot AS$, hence $\tfrac{AF}{AB}=\tfrac{AE}{AC}$ i.e $EF$ $\parallel$ $BC$. Note that circle $(AEF)$ tangent to $\Gamma$ so $PA$ is the tangent line of $(AEF)$. $K$ is the intersection point of $AI$ with $EF$, we get $AP$ passes through the midpoint of $TK$. So by Menelaus theorem for $\triangle SKD$ we imply $\tfrac{PD}{PT}$ $=$ $\tfrac{AD}{AK}$ $=$ $\tfrac{ID}{IK}$. Thus, $PI$ $\parallel EF$ so $\angle AIP$ $=$ $\angle AKT$ $=$ $\angle IAP$ which imply $\triangle IPA$ is isosceles i.e $PA$ $=$ $PI$. $\square$

Actually, I proposed a similar problem in August 2015, in problem weekly, see https://www.dropbox.com/s/crx5o1dyt2mmhca/derakynay1187.pdf

buratinogigle wrote: Actually, I proposed a similar problem in August 2015, in problem weekly, see https://www.dropbox.com/s/crx5o1dyt2mmhca/derakynay1187.pdf Please make files available. Thanks.

2017 Olympic Revenge P2 wrote: Let $\triangle$$ABC$ a triangle with circumcircle $\Gamma$. Suppose there exist points $R$ and $S$ on sides $AB$ and $AC$, respectively, such that $BR=RS=SC$. A tangent line through $A$ to $\Gamma$ meet the line $RS$ at $P$. Let $I$ the incenter of triangle $\triangle$$ARS$. Prove that $PA=PI$ Solution:- Let $RI\cap AC=X$ and $SI\cap AB=Y$ and $AI\cap RS=Z$. Just notice that $SY\|RC$ and $XR\|BS$. So, $$\begin{cases} \frac{AR}{RB}=\frac{AX}{XS} \\ \frac{AY}{AR}=\frac{AS}{AC}\end{cases}\implies \frac{AY}{AB}=\frac{AX}{AC}\implies YX\|BC$$. Let $RS\cap YX=T$ and $AI\cap XY=K$ and $AP\cap XY=J$. So, $$-1=(R,S;Z,T)\overset{A}{=}(Y,X;K,T)$$Also as $\odot(AYX)$ and $\odot(ABC)$ are homothetic we get that $AP$ is tangent to $\odot(AYX)$. So, $JY.JX=JA^2\implies JK=JA$ as $(Y,X;K,T)=-1$. Now by Menelaus we get $\frac{ZA}{AK}=\frac{PZ}{PT}$. Also $(I,A;K,Z)=-1\implies \frac{ZA}{KA}=\frac{ZI}{KI}=\frac{ZP}{PT}\implies IP\|KJ$. So as $AJ=JK$ we get that $PA=PI$. $\blacksquare$.

Let $Q$ on the $RS$ such that $IQ||BC$ , It's enough to say that $P=Q$ . Note that if $K$ is the intersection of $BS,RC$ then it's easy to see that $IRKS$ is parallelogram . $\frac{QR}{QS}=\frac{IR}{IS}\times{\frac{KC}{KB}}=\frac{sin\angle{ACR}}{sin\angle{BCR}}\times{\frac{sin\angle{CBS}}{sin\angle{ABS}}}=\frac{sin\angle{CAK}}{sin\angle{BAK}}=\frac{CS.AR}{SA.RB}\times{\frac{AB}{AC}}=\frac{PR}{PS}$ $\implies$ $P=Q$ and we are done $\blacksquare$

Cute! LittleGlequius wrote: Let $\triangle$$ABC$ a triangle with circumcircle $\Gamma$. Suppose there exist points $R$ and $S$ on sides $AB$ and $AC$, respectively, such that $BR=RS=SC$. A tangent line through $A$ to $\Gamma$ meet the line $RS$ at $P$. Let $I$ the incenter of triangle $\triangle$$ARS$. Prove that $PA=PI$ Let $A'$ be the reflection of $A$ in $\triangle ARS$ and $O$ be its circumcentre. Let bisectors of $\angle R$ and $\angle S$ meet $AS$ and $AR$ at $D$ and $E$ respectively. Note that the problem is equivalent to $\odot(AIA')$ and $\odot(ABC)$ being orthogonal. Invert at $A$ with radius $\sqrt{AR \cdot AS}$ followed by reflection in $AI$, we see that $\{D, B\}, \{E, C\}, \{A', O\}$ swap and $I$ maps to the $A$-excenter $J$ of $\triangle ARS$. Thus we require $JO \perp DE$, which is well-known.

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With $RE \parallel BS$ and $SF \parallel CR$ by Pappus we have $EF \parallel BC$. With $(T,K;E,F)=-1$ and $MA^2=MF\cdot ME$ we have $MK=MT=MA$. Then $PT/PD=AK/AD=IK/ID$ (first by Menelaus on $\triangle DKT$, second by harmonic division). Hence $PI\parallel EF \implies PI/PA=MK/MA=1$. This problem also here and here.

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