Let (O) be circumcircle of the $\triangle ABC.$ Let the circumcircles (U), (V) of the $\triangle ABC_{1},\ \triangle AB_{1}C$ meet BC again at F, E. Let PE, PF meet (U), (V) again at X, Y. Let AP meet BC at K and (O) at Q. $EB_{1},\ FC_{1}$ are antiparallel to AB, AC with respect to the $\angle B,\ \angle C,$ respectively, so that $EB_{1}\perp OB,\ FC_{1}\perp OC.$ From (U), $\angle BXE \equiv \angle BXP = \angle BAP \equiv \angle B_{1}AP$ and from (V), $\angle B_{1}AP = 180^\circ-\angle B_{1}EP = \angle B_{1}EX.$ It follows that $BX \parallel EB_{1}$ and similarly, $CY \parallel FC_{1}.$ Consequently, $BX \perp OB,\ CY \perp OC$ are tangents of (O) at B, C. From (U), (V), $\angle XC_{1}A = \angle XPA \equiv \angle EPA = \angle ECA,$ hence $XC_{1}\parallel BC$ and similarly, $YB_{1}\parallel BC.$ Thus $X,\ B_{1},\ C_{1},\ Y$ are collinear and $XY \parallel BC.$ Consequently, BX = CY. Let the tangents BX, CY of (O) meet at L. Then powers of L to (U), (V) are equal, $LB \cdot LX = LC \cdot LY,$ so that L is on the radical axis AP of (U), (V) and $AP \equiv AL \equiv AK$ is the A-symmedian of the $\triangle ABC.$ This can also be proved by a direct calculation of powers of the A-symmedian foot to (U), (V). From (O), (V), $\angle QBC = \angle QAC \equiv \angle PAC = \angle PEC \equiv \angle PEF$ and similarly from (O), (U) $\angle QCB = \angle PFE.$ Thus $\triangle PEF \sim \triangle QBC$ are similar with coefficient $\frac{EF}{a}.$ Powers of B, C to (V), (U) are $BE \cdot a = \frac{c^{2}}{2},\ CF \cdot a = \frac{b^{2}}{2},$ yielding $\frac{EF}{a}= 1-\frac{BE+CF}{a}= 1-\frac{b^{2}+c^{2}}{2a^{2}}$ By cosine law for the $\triangle ABK$ or $\triangle ACK,$ the A-symmedian length is $AK = \frac{bc}{b^{2}+c^{2}}\ \sqrt{2(b^{2}+c^{2})-a^{2}}$ K divides BC in the ratio $\frac{KB}{KC}= \frac{c^{2}}{b^{2}},$ yielding KB, KC. Power of K to (O) is $KB \cdot KC = AK \cdot KQ,$ yielding $\frac{AK}{KQ}= \frac{AK^{2}}{KB \cdot KC}= 2 \frac{b^{2}+c^{2}}{a^{2}}-1$ As a result, $\frac{AP}{PQ}= \frac{AK+KP}{KQ-KP}= \frac{AK+\frac{KQ \cdot EF}{a}}{KQ-\frac{KQ \cdot EF}{a}}=$ $= \frac{\frac{AK}{KQ}+\frac{ER}{a}}{1-\frac{EF}{a}}= \frac{2 \frac{b^{2}+c^{2}}{a^{2}}-1+1-\frac{b^{2}+c^{2}}{2a^{2}}}{1-1+\frac{b^{2}+c^{2}}{2a^{2}}}= 3$ $AQ = 2AP_{1},$ because the triangles $\triangle ABC \sim \triangle AB_{1}C_{1}$ are similar with center A and coefficient 2, so that $\frac{AP}{AP_{1}}= \frac{2AP}{AQ}= \frac{2AP}{AP+PQ}= \frac{6PQ}{4PQ}= \frac{3}{2}$

Consider inversion with center $A$ and coeficient $\sqrt{\frac{AB\cdot AC}{2}}$ combinated with reflection through the axis of angle $\angle BAC$. Points $B$ and $C_1$ and poitns $C$ and $B_1$ will change places. $P'$ will be point, where $BC_1$ and $CB_1$ meet, therefore it'll be centroid of $\triangle ABC$. $Q'$ will be the point, where $AP'$ and $BC$ meet, therefore it'll be midpoint of $BC$. Therefore $\frac{AQ}{AP}=\frac{AP'}{AQ'}=\frac{2}{3}$, which gives what we wanted. Q.E.D.