We can easily angle chase that $AP=AB$ and $AQ = AC$. Clearly, $CI \parallel RQ$ and $BI \parallel RP$. Since $PQ \parallel BC$, triangles $PRQ$ and $BIC$ are homothetic. Let $D \equiv AI \cap BC$. Notice that $\frac{AP}{AQ} = \frac{AB}{AC} = \frac{DB}{DC}$ so $A$ and $D$ are corresponding points in homothetic triangles $PRQ$ and $BIC$. By angle bisector theorem $\frac{AI}{ID} = \frac{AB}{BD}$. (1) Since $PAR \sim BDI$ we have $\frac{AR}{ID} = \frac{AP}{BD}$. (2) From (1) and (2) it follows that $AI = AR$ and we are done.

Another way to finish it: Once you know that $AP = AB$, $AC = AQ$, $BI \parallel PR$, $CI \parallel QR$, notice that the perpendicular bisectors of $PB$ and $CQ$ meet at the midpoint of $IR$, but $A$ belongs to both perpendicular bisectors, hence $A$ is the midpoint of $IR$ and in particular $AI=AR$.

Let $P' = \overleftrightarrow{BI} \cap \overleftrightarrow{PQ}$ and $Q' = \overleftrightarrow{CI} \cap \overleftrightarrow{PQ}$. It is easy to show that $A$ is midpoint of $PP'$ and $QQ'$. Also, $BI \parallel PR $ and $CI \parallel QR$. If $S_A$ denotes the symmetry with respect to $A$: $S_A \left( \overleftrightarrow{BI} \right) = \overleftrightarrow{PR}$ and $S_A \left( \overleftrightarrow{CI} \right) = \overleftrightarrow{QR}$, then $ S_A \left( \overleftrightarrow{BI} \cap \overleftrightarrow{CI} \right) = \overleftrightarrow{PR}\cap\overleftrightarrow{QR}$, ie, $S_A(I)=R$. Therefore $AI=AR$.

Let $U,V$ be the respective points where the internal bisectors of angles $C,B$ intersect $PR,QR$. Clearly $IU$ is perpendicular to the external bisector of $C$, hence parallel to line $QR=VR$. Simillarly $IV$ is parallel to $PR=UR$, or $IURV$ is indeed a parallelogram. Denote by $I_a,I_b,I_c$ the excenters of $ABC$, or lines $I_aI_b,I_aI_c$ are respectively equal to lines $CQ,BP$. Thus triangles $I_aBI_b,I_aCI_c,UPI_c,VQI_b$ are all similar. Now $PQ$ and $I_bI_c$ intersect at $A$, or $A\in UV$. $PQR$ is clearly the result of performing a homothety to $BCI$ with center $I_a$ because their respective sides are parallel, and $BP,CQ$ meet at $I_a$. Or $R$ is on $II_a$ which is the internal bisector of angle $A$, hence $A\in IR$. We conclude that $A$ is the point where the diagonals $UV,IR$ of parallelogram $IURV$ meet, hence the midpoint of both diagonals. It follows that $IA=AR$.

Let $BP$ cut $CQ$ at $D$. It is well known that $A, I, D$ are collinear. Notice that $(A,E;I,D)$ is harmonic, thus $\frac{DA}{DE} = \frac{IA}{IE}$ As $\Delta PRQ\sim \Delta BIC, \frac{BI}{PR} = \frac{IC}{RQ}$, this means that $R, I, D$ are collinear. Hence $R, A, I, D$ are collinear and $\frac{RA}{IE} = \frac{PA}{BE} = \frac{AD}{ED} = \frac{AI}{IE}$ Thus $RA = AI$

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PP. Let $\triangle ABC$ with the incenter $I$ and $P$ and $Q$ the intersections of the parallel to $BC$ that passes through $A$ with the external bisectors of $\widehat{ABC}$ and $\widehat{ACB}$ respectively. Let $R$ such that $RP\perp BP$ and $RQ\perp CQ$ . Show that $AI = AR$ . Proof. Observe that $I_a\in PB\cap QC$ . Denote $D\in II_a\cap BC$ . Prove easilly that $IB\parallel RP$ , $IC\parallel RQ$ and $\boxed{\triangle PRQ\sim \triangle BIC}\ (*)$ . Therefore, the points $\{R,A,I,D,I_a\}$ are collinearly (using the Desarques' theorem). Since $\frac {AP}{AQ}=\frac cb=\frac {DB}{DC}$ obtain that the points $A\in PQ$ and $D\in BC$ are homologously in the similarity $(*)$ . In conclusion, $\frac {RA}{ID}=\frac {PQ}{BC}=$ $\frac {b+c}{a}=\frac {IA}{ID}$ (using the Van Aubel's relation), from where obtain $RA=IA$ .

I tried solving this for fun! My approach was similar to hatchguy's. First, notice that $BI\bot BP$, since $BI$ and $BP$ are the internal and external angle bisectors of $\angle ABC$, respectively. Since $PR\bot BP$, we have $BI\parallel PR$; similarly, $CI\parallel QR$. Let $\ell_1$ be the parallel to $AB$ through $P$, $\ell_2$ the parallel to $AC$ through $Q$, and $S=\ell_1\cap\ell_2$. We have $PS\parallel AB$ and $QS\parallel AC$, which means $\triangle ABC$ and $\triangle SPQ$ are homothetic. Let $T$ be the point in $PB$ such that $AT$ is the angle bisector of $\angle PAB$. Observe that $\angle PAB\cong\angle ABC\Rightarrow\frac{\angle PAB}{2}=\frac{\angle ABC}{2}$. Therefore, $\angle PAT\cong\angle IBC$, implying $AT\parallel BI$. We then have $AT\bot PB$. This means $AT$ is both an angle bisector and an altitude of $\triangle PAB$; in other words: $\triangle PAB$ is isoceles, $T$ is the midpoint of $PB$, and $AP=AB$. Same argument implies $AQ=AC$. Now I claim $R, A,I$ are collinear. It's probably trivial because of the homotheties, but I wanted to be extra rigorous. Anyways, observe that $\frac{PS}{AP}=\frac{PS}{AB}$ and $\frac{QS}{AQ}=\frac{QS}{AC}$ because of the previous side equalities. But $\triangle ABC\sim\triangle SPQ\Rightarrow\frac{PS}{AB}=\frac{QS}{AC}$; then $\frac{PS}{AP}=\frac{QS}{AQ}$, implying that $AS$ is the angle bisector of $\angle PSQ$. Also, $R$ is clearly the incenter of $\triangle SPQ$, so $R\in AS$. Next, we have $AS\parallel AI$, since they're angle bisectors of corresponding angles of homothetic triangles. It follows that $S,R,A,I$ are collinear. Finally, observe that $\Box BPRI$ is a trapezoid with $BI\parallel AT\parallel PR$. Since $T$ is the midpoint of $BP$, $AT$ must be the median of the trapezoid; thus $AR=AI$, as desired.

From $\triangle PAB$ and $\triangle ACQ$ isosceles it suffices to prove that $RAI$ is collinear. But since $PRQ$ and $BIC$ is homothetic then $RI, PB, QC$ are concurrent at point $I_A$. But $AI, PB, QC$ are concurrent at point $I_A$. So $RAI$ are collinear and we win.

Let $I_A$ be the A-excenter and $\angle B=\beta$ $\angle C=\gamma$ $\angle A=\alpha$ $Claim 1$ $I_A-I-A-R$ collinear We have $I_APRQ$ and $I_ABIC$ cyclic From $PQ\parallel BC$ we have $\angle RPQ= 90-\angle APB= 90-\angle PBA=\frac{\beta}{2}$ From the two cyclism we have $\angle RI_AQ=\angle RPQ=\angle IBC=\angle II_AC$ $\implies$ $I_A-I-A-R$ collinear Now let $RI_A\cap BC=X$ $Claim 2$ $\frac{AR}{XI}=\frac{AI}{XI}$ $\angle PBI=90$ $\implies$ $PR\parallel BI$ We clearly see that $\triangle PAR\sim \triangle BXI$ $\implies$ $\frac{AR}{XI}=\frac{AP}{XB}$ but we also have $\angle APB=\angle ABP$ $\implies$ $AP=AB$ Now we need to prove $\frac{AB}{BX}=\frac{AI}{XI}$ But we have $\angle ABI=\angle IBX=\frac{\beta}{2}$ and $\angle AIB=90+\frac{\gamma}{2}$, $\angle BIX=90-\frac{\gamma}{2}$ thus using Law of Sines in $\triangle AIB$ and $\triangle BIX$ respectively we obtain $\frac{AB}{BX}=\frac{AI}{XI}$ $\implies$ $\frac{AR}{XI}=\frac{AI}{XI}$ $\implies$ $$AP=AI$$

Using the notation as below. We first see that there is a homothety $h$ centered at $I_{A}$ taking $B$ to $P$ and $C$ to $Q$ as $PQ\parallel BC$ Now, we see that since $IB\parallel RP$ and $IC\parallel RQ$, it implies $h(I)=R$ and so we also get $I_A-I-A-R$ collinear. And then $h(A)=A'$ which is also $II_{A}\cap BC$ . Then we let $I'$ be the reflection of $I$ over $A'$. Thus to prove that $A$ is the midpoint of $RI$, it is equivalent to prove that $h(I')=I$ which is also equivalent to prove $\frac{I_{A}A}{I_{A}A'}=\frac{I_{A}I}{I_{A}I'}$ Which is true by some length chasing as these are just known lengths.

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Let $I$ and $I_A$ be the incenter and $A-$excenter of $\triangle ABC$. Note that $IB\perp BP$ and $IC\perp CQ$. Also, since $I_A=BP\cap CQ$, this implies that the homothety with center $I_A$ sending $BC$ to $PQ$ sends $I$ to $R$, so $R,A,I,I_A$ are collinear. Now, let $\infty$ be the point at infinity of line $BC$ and $P'=BI\cap PQ$. Then, \[ -1=B(AC;PI)=(A\infty;PP'), \]so $A$ is the midpoint of $PP'$. But then, since $PR\parallel IP'$ this implies $PRP'I$ is a parallelogram, so $A$ is the midpoint of $RI$ implying the desired result. $\blacksquare$

Let $I_A$ be the $A$-excenter of $\triangle ABC$. Observe that $RPI_AQ$ is cyclic and $R,A,I,I_A$ are collinear. Also $AB=AP$ and $AC=AQ$ due to trivial angle chase. Hence, by PoP, $$RA\cdot AI_A=AQ\cdot AP=AB\cdot AC=AI\cdot AI_A\implies AI=AR,$$as desired.

1. $A$, $I$, and $R$ are collinear. Let $\angle{}CAB=2a$, $\angle{}ABC=2b$, and $\angle{}BCA=2c$. Through basic angle chasing on $R$, we find that \[\angle{}RPA=\angle{}RPB-\angle{}APB=90-(90-b)=b\]Similarly, we also have $\angle{}AQR=c$. We can prove this using phantom points. Let $R'$ the point on line $AI$ such that $\angle{}R'PA=b$. We now claim that $\angle{}AQR'=c$. WLOG, let $BC=sin2a$, the other side lengths follow by Law of Sines. Notice that $\triangle{}APB$ is isosceles, so $AP=AB=\sin{}2c$, and through LoS, it follows that $AR=\sin{}2c*\frac{\sin{}b}{\sin{}a+b}$, and since we know that $\angle{}R'AQ=180-(a+2c)$ and $AQ=AC=\sin{}2b$, we have that $\angle{}AQR'=c$ by SAS, meaning that $R'=R$. 2. $AR=AI$. By Law of Sines on $\triangle{}AIC$, we have that \[AI=AC*\frac{\sin{}c}{\sin{}(180-(a+c))}=AC*\frac{\sin{}c}{\sin{}(a+c)}\]and by Law of Sines on $\triangle{}RAQ$, we have that \[AR=AQ*\frac{\sin{}c}{\sin{}(a+c)}=AC*\frac{\sin{}c}{\sin{}(a+c)}=AI\]and we are done.

Because of the internal and external angle bisectors we'll have that $PQ//BC, RP//IB$ and $RQ//IC \implies \triangle BIC \sim \triangle PRQ \implies$ $R,A,I$ are collinear. Now let $K=BI \cap PQ$ , we have $\angle IBC=\angle RPQ$ so $\angle APB=90^{\circ}-\angle IBC$ but $APB=\frac{180^{\circ}-\angle ABC}{2}=90^{\circ}-\angle IBC \implies \angle APB=\angle ABP$ so $\triangle APB$ is isosceles. $PQ//BC \implies PK//BC \implies \angle PKB=\angle IBC$ and notice $\angle ABI=\angle ABK=\angle IBC$ so $\angle ABK=\angle AKB \implies \triangle ABK $ is also isosceles. Now with this we have $PA=BA=KA$ in fact $A$ is the circumcenter of $\triangle PBK$. Now $\angle RPQ=\angle RPK=\angle AKB=\angle AKI=\angle IBC$ and also $\angle IAK=\angle RAP$ so $\triangle AKI \sim \triangle APR$ but with $AP=AK \implies$ $\triangle APR\equiv\triangle AKI(A.S.A)$ hence $AR=AI$ $\square$