$P,Q$ are obviously isogonal conjugates WRT $\triangle ABC.$ Let $AP,AQ$ cut $BC$ at $N,M.$ Since $\angle BPN =180^{\circ}-\angle APB$ and $\angle CQM=180^{\circ}-\angle AQC,$ the problem is equivalent to prove that there exists a point $D$ on $\overline{BC}$ such that $PN,PD$ are isogonal WRT $\angle BPC$ and $QM,QD$ are isogonal WRT $\angle BQC.$ Assume there exists such a $D.$ By Steiner theorem we have: $ \frac{BN}{NC} \cdot \frac{BD}{DC}=\frac{BP^2}{CP^2} \ , \ \frac{BM}{MC} \cdot \frac{BD}{DC}=\frac{BQ^2}{CQ^2} \ \Longrightarrow$ $\frac{BQ^2}{CQ^2} \cdot \frac{MC}{BM}=\frac{BP^2}{CP^2} \cdot \frac{NC}{BN} \Longrightarrow \ \frac{BQ}{CQ} \cdot \frac{\sin \widehat{AQC}}{\sin \widehat{AQB}}=\frac{BP}{CP} \cdot \frac{ \sin \widehat{APC}}{\sin \widehat{APB}}$ $\Longrightarrow \ \frac{ \sin \widehat{AQC}}{\sin \widehat{QCA}} \cdot \frac{\sin \widehat{QBA}}{\sin \widehat{AQB}}=\frac{\sin \widehat{APC}}{\sin \widehat{PCA}} \cdot \frac{\sin \widehat{PBA}}{\sin \widehat{APB}}$ $\Longrightarrow \ \frac{AC}{AQ} \cdot \frac{AQ}{AB}=\frac{AC}{AP} \cdot \frac{AP}{AB}, $ which is obvious identity. Thus, there exists such a $D.$

Obviously $P,Q$ are isogonal conjugates. Since the second condition is equivalent to $\angle{AQB}+\angle{DQC}=180^\circ$, which is analogous to the first condition (i.e. $P,Q$ are switched), it suffices to prove that the first condition implies the second. By the Law of Sines on $\triangle{DPB},\triangle{DPC}$ in addition to the angle conditions, we have \[\frac{BD}{\sin{APC}}=\frac{BD}{\sin(180^\circ-DPB)}=\frac{BD}{\sin{DPB}}=\frac{DP}{\sin{PBC}},\\ \frac{CD}{\sin{APB}}=\frac{CD}{\sin(180^\circ-DPC)}=\frac{CD}{\sin{DPC}}=\frac{DP}{\sin{PCB}},\]so \[\frac{BD\sin{APB}}{CD\sin{APC}}=\frac{\sin{PCB}}{\sin{PBC}}\qquad(1).\]Now by the Law of Sines on $\triangle{AQB},\triangle{AQC}$ and isogonal conjugates $P,Q$, \[\frac{AB}{\sin{AQB}}=\frac{AQ}{\sin{QBA}}=\frac{AQ}{\sin{PBC}}\\ \frac{AC}{\sin{AQC}}=\frac{AQ}{\sin{QCA}}=\frac{AQ}{\sin{PCB}},\]so from (1), \[\frac{AB\sin{AQC}}{AC\sin{AQB}}=\frac{\sin{PCB}}{\sin{PBC}}=\frac{BD\sin{APB}}{CD\sin{APC}}\qquad(2).\]But again by the Law of Sines on $\triangle{DQB},\triangle{DQC}$ and isogonal conjugates $P,Q$, we have \[\frac{BD}{\sin{DQB}}=\frac{DQ}{\sin{QBC}}=\frac{DQ}{\sin{PBA}},\\ \frac{CD}{\sin{DQC}}=\frac{DQ}{\sin{QCB}}=\frac{DQ}{\sin{PCA}},\]so dividing and applying the Law of Sines on $\triangle{PCA},\triangle{PBA}$, \[\frac{BD\sin{DQC}}{CD\sin{DQB}}=\frac{\sin{PCA}}{\sin{PBA}}=\frac{\frac{AP}{AC}\sin{APC}}{\frac{AP}{AB}\sin{APB}}=\frac{AB\sin{APC}}{AC\sin{APB}}.\]Multiplying this equation with (2), we obtain \[\sin{DQC}\sin{AQC}=\sin{DQB}\sin{AQB},\]or (note that $(DQC+AQC)=360^\circ-(DQB+AQB)$) \begin{align*} \cos(DQC-AQC)-\cos(DQC+AQC)&=\cos(DQB-AQB)-\cos(DQB+AQB)\\ \implies\cos(DQC-AQC)&=\cos(DQB-AQB).\end{align*}Clearly angles $DQC,AQC,DQB,AQB$ are all in $(0,180^\circ)$, so their differences are in $(-180^\circ,180^\circ)$. Thus we either have \[DQC-AQC=DQB-AQB \implies DQC+AQB=DQB+AQC=180^\circ,\]as desired, or \[DQC-AQC=-DQB+AQB \implies DQC+DQB=AQC+AQB=180^\circ,\]contradicting the fact that $Q$ is inside $\triangle{ABC}$.

All angles in this proof are directed. Lemma 1: Let $Q$ be an interior point of $\triangle ABC$, and let $D$, $E$, and $F$ be the projections of $Q$ onto $BC$, $CA$, and $AB$, respectively. Let $O$ be the circumcenter of $\triangle DEF$, and let $M$ be the midpoint of $OA$. Then $\angle MOF = \angle ACQ + \angle QBA.$ Proof: $FE$ is a common chord of the circumcircles of $AFQE$ and $\triangle DEF$, so $MO \perp EF$. Hence, \[ \angle MOF \&= \frac{\angle EOF}{2} \&= \angle EDF = \angle EDQ + \angle QDF \&= \angle ACQ + \angle QBA. \] Corollary: Let $P'$ be the reflection of $Q$ across $O$, and let $P$ be the isogonal conjugate of $Q$. Then $P' = P$. Proof: Let $M$ and $N$ be the midpoints of $AQ$ and $BQ$, respectively, and let $F$ be the projection from $Q$ onto $AB$. By the lemma, \begin{align*} \angle MON &= \angle MOF + \angle NOF = (\angle ACQ + \angle QBA) + (\angle QCB + \angle BAQ) \\ &= (\angle PCB + \angle PBC) + (\angle PCA + \angle CAP) \\ &= (180^{\circ} - \angle BPC) + (180^{\circ} - \angle CPA) \\ &= \angle APB. \end{align*} A homothety centered at $Q$ with factor 2 reveals that $\angle AP'B = \angle APB$. Similarly, we find that $\angle CP'A = \angle CPA$ and $\angle BP'C = \angle BPC$. Hence, $P'$ lies on the intersection of the circumcircles of $\triangle ABP$, $\triangle CPA$, and $\triangle BPC$. But these circumcircles intersect only at the point $P$, so we have $P = P'$, as desired. Let $R$ and $S$ be the reflections of $P$ and $Q$ across $BC$, respectively, and let $D' = PS \cap QR$. Because $D'$ lies on the axis of symmetry of isosceles trapezoid $PRSQ$, $D'$ must lie on $BC$ as well. As $D$ varies from $C$ to $B$, $\angle DPC$ increases and $\angle DQB$ decreases, so there exists at most one point $D$ such that $\angle APB + \angle DPC = 180^{\circ}$ and $\angle AQC + \angle DQB = 180^{\circ}$. We claim that $D = D'$ satisfies both of these conditions. By symmetry, it is sufficient to show that $\angle APB + \angle D'PC = 180^{\circ}$. Let $O$ be the the midpoint of $PQ$, let $M$ be the midpoint of $CQ$, and let $T$ be the midpoint of $QS$. By the corollary to our lemma, $O$ is the circumcenter of the $Q$-pedal triangle of $\triangle ABC$, so by the lemma, $\angle TOM = \angle QCA + \angle ABQ = 180^{\circ} - \angle APB$. A homothety centered at $Q$ with factor 2 sends $M$ to $C$, $O$ to $P$, and $T$ to $S$, so $\angle TOM = \angle SPC = \angle D'PC = 180^{\circ} - \angle APB$, which completes our proof.

Angle chasing yields that the isogonal conjugate of $A$ with respect to $\triangle{BPC}$ is the reflection $Q'$ of $Q$ in line $BC$. The condition $\angle APB+\angle DPC = 180^\circ$ is therefore equivalent to $P$, $Q'$ and $D$ being collinear. Similarly, the condition $\angle AQC+\angle DQB = 180^\circ$ is equivalent to $Q$, $P'$ and $D$ being collienear where $P'$ is the reflection of $P$ in $BC$. Now note that quadrilaterals $BPCQ'$ and $BP'CQ$ are reflections in $BC$. Therefore $PQ'$ and $P'Q$ divide $BC$ in the same ratio and concur on $BC$. This implies the result.

Let $ AP \cap BC=S $ and $ AQ \cap BC=T $ and $ \angle BAP=\angle CAQ=x, \angle CBP=\angle ABQ=y $ and $ \angle ACQ=\angle BCP=z $.After Steiner's theorem , this problem is equivalent to $ \frac{BP^2}{CP^2} \cdot \frac{BS}{SC}=\frac{BQ^2}{CQ^2} \cdot \frac{BT}{ST} (*) $.By sine law to the triangles $ BPC $ and $ BQC $, $ (*) $ is equivalent to $ ( \frac{sinx}{sin(A-x)} \cdot \frac{siny}{sin(B-y)} \cdot \frac{sinz}{sin(C-z)})^2=1 $, which is true by Ceva's theorem.

My solution: Lemma : Let $ P,P' $ be the isogonal conjugate of $ \triangle ABC $ . Let $ \ell $ be the isogonal conjugate of $ AP $ of $ \angle BPC $ . Let $ \ell ' $ be the isogonal conjugate of $ AP' $ of $ \angle BP'C $ . Then $ \ell, \ell' $ are symmetry WRT $ BC $. Proof of the lemma: Let $ X $ be the intersection of $ BP' $ and $ CP $ . Let $ Y $ lie on $ BC $ satisfy $ XA,XY $ are isogonal conjugate of $ \angle BXC $ . Since $ CA,CY $ are isogonal conjugate of $ \angle P'CP $ , so $ A,Y $ are isogonal conjugate of $ \triangle CXP' $ . ie. $ P'Y,P'A $ are isogonal conjugate of $ \angle BP'C $ . Similarly, we can prove $ PY, PA $ are isogonal conjugate of $ \angle BPC $ . Since $ \angle PYB=\angle CYP' $ (Easy angle chasing) , so we get $ PY,P'Y $ are symmetry WRT $ BC $ . Back to the main problem : From lemma we get $ \angle APB+\angle DPC=180^{\circ} \Longleftrightarrow PA, PD $ are isogonal conjugate of $ \angle BPC $ $ \Longleftrightarrow QA, QD $ are isogonal conjugate of $ \angle BQC \Longleftrightarrow \angle AQC+\angle DQB=180^{\circ} $ Q.E.D

Let $\triangle A'B'C'$ be the pedal triangle of $Q$ WRT $\triangle ABC$ and let $\triangle Q_aQ_bQ_c$ be the triangle obtained by reflecting $Q$ in $BC, CA, AB.$ Since $P, Q$ are isogonal conjugates, it is well-known (Corollary to Fact 3) that $P$ is the circumcenter of $\triangle Q_aQ_bQ_c.$ Therefore $BP$ is the perpendicular bisector of $\overline{Q_aQ_c}$, implying that \[\measuredangle BQ_aP = -\measuredangle Q_aPB - \measuredangle PBQ_a = -\measuredangle Q_aPB - \measuredangle ABC = -\measuredangle Q_aQ_bQ_c - \measuredangle ABC.\]Furthermore, since $Q, A, B', C'$ are inscribed in the circle of diameter $\overline{AQ}$, with similar relations holding for the two other sets of vertices, we have \[\measuredangle Q_aQ_bQ_c = \measuredangle A'B'C' = \measuredangle A'B'Q + \measuredangle QB'C' = \measuredangle A'CQ + \measuredangle QAC' = \measuredangle BCQ + \measuredangle QAB = -\measuredangle CQA - \measuredangle ABC,\]where the last step follows from examining quadrilateral $ABCQ.$ Consequently, if $Q_aP$ cuts $BC$ at $D'$, then \[\measuredangle BQD' = -\measuredangle BQ_aD' = -\measuredangle BQ_aP = -\measuredangle CQA,\]implying that $D' \equiv D.$ Thus, if $P_a$ is the reflection of $P$ in $BC$, then by symmetry in trapezoid $QQ_aP_aP$, we have $D \in QP_a$, and analagous arguments show that $\measuredangle CPD = \measuredangle APB.$ The desired result follows. $\square$

For uniqueness reasons it suffices to show that there is a single point $D$ satisfying both conditions at once. To construct it, take the ellipse with foci $P$ and $Q$ tangent to the sides $BC$, $CA$, $AB$ at $D$, $E$, $F$. We claim this works. It's a known property of ellipses that $\overline{CP}$ bisects $\angle DPE$. Similarly, $\overline{BP}$ bisects $\angle FPD$ and $\overline{AP}$ bisects $\angle EPF$. Thus $\angle APB + \angle DQC = 180^{\circ}$ follows, and in the same way $\angle AQC + \angle DQB = 180^{\circ}$.

My solution Lemma: If the points $P^\ast, Q^\ast $ are symmetric to the points $ P, Q $ with respect to the line $ BC $, then $ P^\ast$ and $ A $ are isogonal conjugates with respect to $\triangle BQC,$ and also $Q^\ast$ and $ A $ are isogonal conjugates with respect to $\triangle BPC.$ [asy][asy] import olympiad; size(12cm); defaultpen(fontsize(10pt)); pair A = dir(120); pair B = dir(210); pair C = dir(330); pair X = (7A+2C)/(7+2); pair Y = (5A+3B)/(5+3); pair P = extension(B,X,C,Y); pair P0 = foot(P,B,C); pair Ps = 2*P0 - P; pair X0 = foot(P,A,C); pair S = 2*X0 - P; pair Y0 = foot(P,A,B); pair K = 2*Y0 - P; pair Q = circumcenter(Ps,S,K); pair Q0 = foot(Q,B,C); pair Qs = 2*Q0 - Q; pair D = extension(P,Qs,Q,Ps); pair T = (8P-5A)/(8-5); pair Z = (5Q-1.5*A)/(5-1.5); draw(A--T, dotted); draw(A--Z, dotted); draw(A--B--C--A,red); draw(B--P--C, black); draw(B--Q--C, black); draw(Q--Ps, dashed); draw(P--Qs, dashed); draw(B--Ps, black); draw(C--Ps, black); draw(B--Qs, black); draw(C--Qs, black); draw(anglemark(Qs,B,C), blue); draw(anglemark(D,B,Q), blue); draw(anglemark(P,B,A), blue); draw(anglemark(A,C,P), darkgreen); draw(anglemark(Q,C,D), darkgreen); draw(anglemark(D,C,Qs), darkgreen); dot("$A$", A, dir(A)); dot("$B$", B, dir(B)); dot("$C$", C, dir(C)); dot("$Q$", Q, dir(120)); dot("$P$", P, dir(120)); dot("$P^\ast$", Ps, dir(Ps)); dot("$Q^\ast$", Qs, dir(Qs)); dot("$D$", D, dir(380)); [/asy][/asy] Note that, $$\measuredangle PBA=\measuredangle CBQ=\measuredangle Q^{\ast}BC,$$$$\measuredangle PCA=\measuredangle BCQ=\measuredangle Q^{\ast}CB,$$therefore $P^\ast$ and $ A $ are isogonal conjugates with respect to $\triangle BQC.$ Similarly, $ Q^\ast $ and $A$ are isogonal conjugates with respect to $\triangle BPC.\blacksquare $ Therefore, the lines $PA$ and $ PQ^{\ast} $ are isogonal conjugates relative to $\angle BPC $, as well as the lines $ QA $ and $ QP^{\ast} $ are isogonal conjugates relative to $\angle BQC, $ whence $$ \angle APB + \angle Q^{\ast}PC = 180^{\circ}, $$$$ \angle AQC + \angle P^{\ast}QB = 180^{\circ}, $$Then it suffices to prove that $ D $ lies on the ray $ QP^{\ast} $ if and only if $D$ lies on the ray $ PQ^{\ast}, $ which is obvious, since $ QP^{\ast} $ and $ PQ^{\ast}$ intersect on $ BC. $

Observe that the point $D_1$ such that $\angle APB+\angle D_1PC=180^\circ$ must be such that $PD_1$ and $AP$ are isogonal conjugates relative to $\angle BPC$. Let $AP$ intersect line $BC$ at $X$, then this result implies \[\frac{BX}{XC}\cdot\frac{BD_1}{D_1C}=\frac{BP^2}{PC^2}.\]Then if we define $Y$ as the intersection of lines $AQ$ and $BC$, it suffices to check \[\frac{BP^2}{PC^2}\cdot\frac{XC}{XB}=\frac{BQ^2}{QC^2}\cdot\frac{YC}{YB}\iff \frac{BP^2}{PC^2}\cdot\frac{XC^2}{XB^2}=\frac{BQ^2}{QC^2}\cdot\frac{YC}{YB}\cdot\frac{XC}{XB}=\frac{BQ^2}{QC^2}\cdot\frac{AC^2}{AB^2}.\]So it is equivalent to check \[1=\frac{BP\cdot XC\cdot QC\cdot AB}{BQ\cdot AC\cdot PC\cdot XB}=\frac{AP\cdot \frac{\sin\angle BAP}{\sin\angle ABP}\cdot AC\cdot\frac{\sin\angle PAC}{\sin\angle AXC}\cdot AQ\cdot\frac{\sin\angle QAC}{\sin\angle QCA}\cdot AB}{AQ\cdot\frac{\sin\angle BAQ}{\sin\angle ABQ}\cdot AC\cdot AP\cdot\frac{\sin\angle PAC}{\sin\angle ACP}\cdot AB\cdot\frac{\sin\angle PAB}{\sin\angle AXB}}=\]\[\frac{\frac{\sin\angle BAP}{\sin\angle ABP}\cdot \frac{1}{\sin\angle PCB}}{\frac{\sin\angle PAC}{\sin\angle PBC}\cdot \frac{1}{\sin\angle ACP}}=\frac{\sin\angle BAP}{\sin\angle ABP}\cdot\frac{\sin\angle PBC}{\sin\angle PCB}\cdot\frac{\sin\angle ACP}{\sin\angle PAC}=\frac{BP}{AP}\cdot\frac{PC}{PB}\cdot\frac{AP}{CP},\]which is trivial.

First, we have a theorem (Isogonal Ratios Theorem): If $AX$ and $AY$ are isogonal with respect to $\angle{BAC}$ and $X, Y \in BC$, then $$\frac{BD\cdot BE}{CD\cdot CE} = \left(\frac{AB}{AC}\right)^2$$ We prove the if version because the only if part is symmetrical. Notice that $P$ and $Q$ are isogonal conjugates. Let $E$ and $F$ be the intersections of $AP$ and $AQ$ with $BC$ respectively. By our theorem, we have the two equations: $$\frac{BE\cdot BF}{CE\cdot CF} = \left(\frac{AB}{AC}\right)^2 \implies \frac{BF}{CF} = \left(\frac{AB}{AC}\right)^2\cdot \frac{CE}{BE}$$$$\frac{BE\cdot BD}{CE\cdot CD} = \left(\frac{BP}{CP}\right)^2 \implies \frac{BD}{CD} = \left(\frac{BP}{CP}\right)^2\cdot \frac{CE}{BE}$$and we wish to prove $$\left(\frac{BQ}{CQ}\right)^2 = \frac{BD\cdot BF}{CD\cdot CF} = \left(\frac{AB\cdot BP\cdot CE}{AC\cdot CP\cdot BE}\right)^2 \implies 1 = \frac{AB\cdot BP\cdot CE\cdot CQ}{AC\cdot CP\cdot BE\cdot BQ}.$$ Now, notice by Law of Sines that $\frac{AB}{BE} = \frac{\sin{\angle{AEB}}}{\sin{\angle{BAE}}}$ and $\frac{AC}{CE} = \frac{\sin{\angle{AEC}}}{\sin{\angle{CAE}}} = \frac{\sin{\angle{AEB}}}{\sin{\angle{CAE}}}$ so $$\frac{AB\cdot BP\cdot CE\cdot CQ}{AC\cdot CP\cdot BE\cdot BQ} = \frac{BP\cdot CQ}{CP\cdot BQ}\cdot \frac{\sin{\angle{CAE}}}{\sin{\angle{BAE}}}.$$ Additionally, $\frac{BP}{CP} = \frac{\sin{\angle{BCP}}}{\sin{\angle{CBP}}}$ and $\frac{CQ}{BQ} = \frac{\sin{\angle{CBQ}}}{\sin{\angle{BCQ}}}$. Therefore, we have $$\frac{BP\cdot CQ}{CP\cdot BQ}\cdot \frac{\sin{\angle{CAE}}}{\sin{\angle{BAE}}} = \frac{\sin{\angle{BCP}}\cdot\sin{\angle{CBQ}}\cdot\sin{\angle{CAE}}}{\sin{\angle{CBP}}\cdot\sin{\angle{BCQ}}\cdot\sin{\angle{BAE}}}$$$$= \frac{\sin{\angle{BCP}}\cdot\sin{\angle{ABP}}\cdot\sin{\angle{CAP}}}{\sin{\angle{CBP}}\cdot\sin{\angle{ACP}}\cdot\sin{\angle{BAP}}} = 1$$by Trig Ceva's, as desired. $\blacksquare$

WLOG, assume $\angle APB + \angle DPC = 180^{\circ}$. In addition, we define $X = AP \cap BC$ and $Y = AQ \cap BC$. It's clear that $P$ and $Q$ are isogonal conjugates wrt $ABC$. Notice $$\angle BPX = 180^{\circ} - \angle APB = \angle DPC$$which implies $PX$ and $PY$ are isogonal in $\angle BPC$. Thus, applying Steiner's Ratio Theorem twice yields $$\frac{AB^2}{AC^2} = \frac{BX}{XC} \cdot \frac{BY}{YC}$$and $$\frac{PB^2}{PC^2} = \frac{BX}{XC} \cdot \frac{BD}{DC}.$$Now, the LoS and Ratio Lemma give $$\frac{QB^2}{QC^2} = \left( \frac{\sin BAQ \cdot \frac{AQ}{\sin ABQ}}{\sin CAQ \cdot \frac{AQ}{\sin ACQ}} \right)^2 = \left( \frac{\sin BAY}{\sin CAY} \cdot \frac{\sin ACQ}{\sin ABQ} \right)^2$$$$= \left(\frac{BY}{YC} \cdot \frac{AC}{AB} \cdot \frac{\sin PCB}{\sin PBC} \right)^2 = \left(\frac{BY}{YC} \cdot \frac{AC}{AB} \right)^2 \cdot \frac{PB^2}{PC^2}$$$$= \left(\frac{AB}{AC} \cdot \frac{XC}{BX} \right)^2 \cdot \frac{BX}{XC} \cdot \frac{BD}{DC} = \frac{AB^2}{AC^2} \cdot \frac{XC}{BX} \cdot \frac{BD}{DC} = \frac{BY}{YC} \cdot \frac{BD}{DC}.$$Hence, the Converse of Steiner's implies $\angle BQD = \angle YQC$, which clearly finishes. $\blacksquare$ Remarks: This problem should be pretty straightforward, as utilizing Steiner's is the obvious approach. We can also reverse engineer certain ratios to help us simplify our trigonometric expression.

Clearly, $P$ and $Q$ are isogonal conjugates. So if we denote $AP\cap BC=\{X\}$ and $AQ\cap BC=\{Y\}$, we have that $\angle BAY=\angle CAX$. So the conclusion would be equivalent with \[\left(\frac{QB}{QC}\right)^2\cdot \frac{YC}{YB}=\left(\frac{BP}{PC}\right)^2\cdot \frac{XC}{XB}\]which reduces, after using the sine law in the triangles $\Delta BPA$, $\Delta BQA$, $\Delta CPA$ and $\Delta QCA$, to \[\frac{\sin{\angle QCA}}{\sin{\angle QCB}}\cdot \frac{\sin{\angle QBC}}{\sin{\angle{QBA}}}\cdot \frac{\sin{\angle BAQ}}{\sin{\angle{CAQ}}}=1\]which is true, by Ceva.

Lemma: Point $P$ inside quadrilateral $ABCD$ has an isogonal conjugate iff $\angle APB + \angle CPD = 180$. Proof: Construct the pedal quadrilateral of P, it's not hard to show that both conditions are equivalent to the pedal quadrilateral being concyclic. Applying this lemma on degenerate quadrilateral $ABDC$ kills the problem.

here lies dumb solution (good solution hopefully coming soon.) Let $AP\cap BC=X$ and let $AQ\cap BC=Y$. It suffices to show that \[\left(\frac{PB}{PC}\right)^2\div \frac{XB}{XC}=\left(\frac{QB}{QC}\right)^2\div \frac{YB}{YC}\]or that \[\frac{PB}{PC}\div \frac{QB}{QC}=\sqrt{\frac{XB}{XC}\div \frac{YB}{YC}}\] Now use Ratio Lemma on $\triangle ABC$ with $AX$, $AY$ as cevians. After some annoying calculations we can find \[\frac{XB}{XC}\div \frac{YB}{YC}=\frac{\sin^2 \angle BAP}{\sin^2 \angle CAP}\]and LoS on $\triangle ABP$ and $\triangle ACP$ gives \[\sin BAP=\frac{PB\cdot \sin \angle ABP}{PA}\]and dividing gives \[\frac{XB}{XC}\div \frac{YB}{YC}=\left(\frac{PB\cdot QC}{PC\cdot QB}\right)^2\]and we're done.

Obviously $P$ and $Q$ are isogonal conjugates. Then we may consider the inellipse with foci $P$ and $Q$, inscribed in $\triangle A BC$and denote by $X$, $Y$ and $Z$ its tangency points to $\overline{BC}$, $\overline{CA}$ and $\overline{AB}$. Claim: The point $X$ is the unique point along $\overline{BC}$ satisfying $\angle APB + \angle XPC = 180$. Proof. The uniqueness part is not hard to see. To show the angle equality note that, \begin{align*} \angle APB + \angle XPC &= (\angle APZ + \angle ZPB) + \angle XPC\\ &= \angle APY + \angle BPZ + \angle CPX\\ &= \frac{1}{2}(\angle YPZ + \angle ZPX + \angle XPY)\\ &= \frac{1}{2}(360) = 180 \end{align*}proving the claim. $\square$ Then an identical angle chase for $Q$ proves the claim and we're done.

Clearly $P$ and $Q$ are isogonal conjugates in $\triangle ABC$. The condition $\angle APB + \angle DPC = 180^\circ$ implies that $P$ has an isogonal conjugate wrt degenerate quadrilateral $ABDC$, which is clearly $Q$ as $\angle ACB = \angle ACD$ and $\angle ABC = \angle ABD$. Clearly the reverse holds true as well, so we are done. This problem also follows from the fact that the inellipse with foci $P$ and $Q$ is tangent to sides $BD$ and $CD$ of deg. quadrilateral $ABDC$, so $P$ and $Q$ are isogonal conjugates in $ABCD$.

Given that \( P \) and \( Q \) are isogonal conjugates with respect to \( \triangle ABC \), we know that \( AP \) and \( AQ \) are isogonal lines, meaning the angles they form with the triangle's sides are equal when measured from corresponding vertices. If \( N \) is the intersection of \( AP \) with \( BC \) and \( M \) is the intersection of \( AQ \) with \( BC \), we aim to prove the existence of a point \( D \) on \( BC \) such that \( PN \) and \( PD \) are isogonal with respect to \( \angle BPC \) and \( QM \) and \( QD \) are isogonal with respect to \( \angle BQC \). To approach this problem, we will use the properties of trigonometric Ceva's theorem and the ratios resulting from the sine rule. First, applying Steiner's theorem to the triangles formed by \( P \), \( Q \), and segments on \( BC \), and specifically using the points \( N \), \( M \), and \( D \), we get the following relationships: \[ \frac{BN}{NC} \cdot \frac{BD}{DC} = \frac{BP^2}{CP^2} \]\[ \frac{BM}{MC} \cdot \frac{BD}{DC} = \frac{BQ^2}{CQ^2} \] By equating the product of these ratios, we find: \[ \frac{BQ^2}{CQ^2} \cdot \frac{MC}{BM} = \frac{BP^2}{CP^2} \cdot \frac{NC}{BN} \] Using the sine rule in triangles \( \triangle BPA \), \( \triangle BQA \), \( \triangle CPA \), and \( \triangle QCA \), we can express these ratios in terms of sines of angles, giving us: \[ \frac{BQ}{CQ} \cdot \frac{\sin \angle AQC}{\sin \angle AQB} = \frac{BP}{CP} \cdot \frac{\sin \angle APC}{\sin \angle APB} \] From the isogonal property, we have: \[ \frac{\sin \angle AQC}{\sin \angle QCA} \cdot \frac{\sin \angle QBA}{\sin \angle AQB} = \frac{\sin \angle APC}{\sin \angle PCA} \cdot \frac{\sin \angle PBA}{\sin \angle APB} \] Simplifying this using the fact that \( P \) and \( Q \) are isogonal conjugates, we obtain: \[ \frac{AC}{AQ} \cdot \frac{AQ}{AB} = \frac{AC}{AP} \cdot \frac{AP}{AB} \] This is an obvious identity as both sides simplify to 1. Thus, there exists a point \( D \) on \( \overline{BC} \) that satisfies the required conditions. By verifying the conditions, we conclude that \( PN \) and \( PD \) are isogonal with respect to \( \angle BPC \), and \( QM \) and \( QD \) are isogonal with respect to \( \angle BQC \). Therefore, the existence of such a point \( D \) is guaranteed by the properties of isogonal conjugates and the applications of trigonometric Ceva's theorem.

Note that $P$ and $Q$ are isogonal conjugates wrt $\Delta ABC$. So $\angle APB + \angle DPC = 180^\circ \iff P$ and $Q$ are isogonal conjugates in $ABDC \iff \angle AQC + \angle DQB = 180^\circ$.

The given angle conditions imply that $P$ and $Q$ are isogonal conjugates in $\Delta ABC$. Then $\angle APB +\angle CPD=180^\circ$ implies $P$ has an isogonal conjugate in the degenerate quadrilateral $ABDC$ so it must in fact be $Q$ so the desired result follows. Same goes for the other direction as they are both symmetric.