Solution

Let $J_A$ be the center of $\omega_A$, $L$ to be on $AP$ such that $LP=LX$, and $K$ be the intersection of $AO$ with $XJ_A$. If $AB=AC$, then the two tangents are parallel, so $R$ does not exist. If $AB\neq AC$, WLOG assume $AB<AC$. As $AB_1=AC_1$ and $AJ_A$ is angle bisector of $\angle B_1AC_1$, $P$ is also on $AJ_A$. Lemma 1. $A$, $K$, $L$, $X$ are concyclic. Proof 1. $\angle LXK=\angle LXP-\angle PXJ_A=\angle LPX-\angle XA_1J_A=\angle AJ_AA_1=\frac{\angle B-\angle C}{2}=\angle PAO=\angle LAK$. Lemma 2. $LK\parallel PX$. Proof 2. As $J_AP\cdot J_AA=J_AC_1^2=J_X^2$, we can see $\angle J_AXP=\angle J_AAX=\angle LAX=\angle LKX$. Theorem. $RP=RX$. Proof. As $\angle RAK=\angle RXK=90^{\circ}$, $R$ is concyclic with the points in Lemma 1, and $RK$ is the diameter. Since $RL$ is perpendicular to $LK$, which is parallel to $PX$, $R$ is on the perpendicular bisector of $PX$, hence $RP=RX$.

Take a homothety sending excircle to mixtilinear circle touching the circumcircle at $T$. Now we want to prove that the pole of $AT$ is on the line through $I$ parallel to $BC$, which is well-known.

Since the polar line of $P$ with respect to $\omega_A$ is $TA$ where $TA$ is $A-$external angle bisector with $T\in BC$, we see that $T,R,X$ are collinear. Also $A,A_1,I_A,X$ are concyclic since $PA.PI_A=PB_1.PC_1=PA_1.PX$. Perform $\sqrt{bc}$ inversion and reflect over the angle bisector of $\measuredangle CAB$. New Problem Statement: $ABC$ is a triangle with $A-$mixtilinear incircle $\omega$. Circumcenter of $\omega$ is $P$ and $A_1$ is the tangency point. Let $I$ be the incenter of $ABC$, $A_1I$ meets $\omega$ at $X$ and $R$ lies on the parallel line from $A$ to $BC$ and $(ARX)$ is tangent to $\omega$. Prove that $\frac{PA}{PR}=\frac{XA}{XR}$. Let $N$ be the midpoint of arc $BAC$. We observe that $A_1,I,X,N$ are collinear. \[\measuredangle (NI,BC)+\measuredangle PXI=180-\measuredangle CNA_1-\frac{\measuredangle B+\measuredangle C}{2}+\measuredangle MAA_1=180-\measuredangle CAM-\frac{\measuredangle B+\measuredangle C}{2}=90\]Thus, $PX\perp BC\parallel AR$. Since $(XAR)$ is tangent to $\omega$, we get that $PX$ is the perpendicular bisector of $AR$ which implies $\frac{PA}{PR}=1=\frac{XA}{XR}$ as desired.$\blacksquare$