Solution: (O2),(O1) are the excircles of triangle APB and APC Because of ∠PAB=∠PBC and ∠PAC=∠PCB so BC is the common tangent of (O1) and (O2). AP is the radical axis of (O1) and (O2) so AP intersecs BC at M where M is the midpoint of BC. So ∠OMQ=90 (1) Triangle QAB and QCA are similar so QA^2=QB*QC. Thus, QA is the tangent of the circle (O) or ∠OAQ=90 (2) From (1) and (2) O,A,Q,M are concyclic So ∠OMA=∠OQA But we also have ∠OMA=∠MQO1 (Q,O1,O2 are collinear) So ∠MQO1=∠AQO. So ∠MQO=∠AQO1. Final ∠AQP=2∠AQO1=2∠OQB.

Here's another approach, slightly different to what Huyen did: I'll use the diagram in the previous post. $ Q, O_{2}, O_{1}$ are collinear since they all lie on the perpendicular bisector of $ AP$. If $ AP$ meets $ BC$ at $ A_{1}$, we have $ BPA_{1}$ and $ ABA_{1}$ are similar; also $ CPA_{1}$ and $ A_{1}CA$ are similar, so $ BA_{1}^2 = AA_{1} * PA_{1} = CA_{1}^2$, so $ BA_{1} = CA_{1}$. Since $ QB, BO_{2}$ are perpendicular, and $ QC, CO_{1}$ are perpendicular, triangles $ QO_{2}B$ and $ QO_{1}C$ are similar, so $ \frac {O_{2} A}{O_{1} A} = \frac {O_{2}B}{O_{1}B} = \frac {QO_{2}}{QO_{1}}$. Considering triangle $ AO_{1}O_{2}$, and the point $ Q$ on $ O_{2} O_{1}$, by the converse of the external version of the Angle Bisector Theorem, $ QA$ is the external angle bisector of $ AO_{2}O_{1}$. A quick angle chase gives: $ \angle O_{2}AB = 90 - B$, $ \angle OAB = 90 - C$, so $ \angle OAO_{2} = A$, and similarly $ \angle O_{1}AO = A$, so $ \angle O_{1}AO_{2} = 2A$, so since $ QA$ bisects the external angle $ O_{2}AO_{1}$ and $ AO$ bisects the internal angle $ O_{2}AO_{1}$, $ \angle QAO = 90$. Thus $ QA$ is tangent to the circumcircle of $ ABC$, and $ QAB$ is similar to $ QAC$. Thus $ QA^2 = QB * QC$ ... (1). Since $ OO_{2}$ is perpendicular to $ AB$, $ OO_{2}$ bisects angle $ AOO_{2}$, so $ \angle AOO_{2} = C$, and since $ \angle OAO_{2} = A$, it follows that $ AO_{2}O$ is similar to $ ABC$. Similarly, $ AOO_{1}$ is similar to $ ABC$, and also $ AOO_{2}$. This gives us: $ AO^2 = AO_{2} * AO_{1}$ ... (2). Combining (1) and (2), gives us: $ (\frac {AQ}{AO})^2 = (\frac {QB}{O_{2}B})(\frac {QC}{O_{1}C})$ ... (3) Since $ \frac {QB}{O_{2}B} = \frac {QC}{O_{1}C}$, since $ QBO_{2}$ and $ QCO_{1}$ are similar, we have $ \frac {AQ}{AO} = \frac {QB}{O_{2}B}$. Since $ \angle QBO_{2} = \angle QAO$, we have $ QBO_{2}$ and $ QAO$ are similar triangles, so $ \angle AQO = \angle O_{2}QB$, so $ \angle AQO_{2} = \angle OQB$, so $ \angle AQP = 2 \angle AQO_{2} = 2\angle OQB$, as required. Some other properties of the point $ P$: - The reflection of $ P$ about the midpoint $ A_{1}$ of $ BC$ lies on the circumcircle. - If $ Q$ is the reflection $ P$ about $ BC$, $ Q$ lies on the circumcircle, and further, $ AQ$ is a symmedian of triangle $ ABC$. - $ \frac {BP}{CP} = \frac {AB}{AC}$, so $ P$ lies on the $ A$- Apollonius circle of $ ABC$.

Another way: suppose R is the reflection of P in BC, it's easy to see that A,B,R,C is cyclic. So OA=OR. since QA=QP=QR, so A,P,R lies on the circle center at Q with radius QA. so <ARP=half of <AQP since OA=OR, QA=QR, so OQ is perpendicular to AR, and since PR is perpendicular to QC, so it's easy to see <ARP=<OQB thus <AQP=s<OQB

Here is another solution:

denote $S$ the midpoint of $AP$,$H$ orthocenter of $\triangle ABC$ ,$M=AP\cap BC$.$A'$ symmetric point of $A$ wrt $M$,. Since $\triangle MPB\sim \triangle MAB$ ie $MB^2=MA*MP$ similarly $MC^2=MP*MA$ so $MB=MC$ now diagonals of $ABA'C$ bisect each other so $ABA'Ç$ is a parallelogram. now $\angle A'CH=90$ but since $\angle APC=180-\angle PAB-\angle PAC=180-\angle ABC=\angle BHC=180-\angle BA'C$ we have $C,P,H,B,A'$ are con-cyclic. so $\angle HPA'=90$ since $QA=QP$ ie $SP=SA=AP/2$ and $\angle QSM=\angle QSP=90$ since $\angle PAH=\angle SQM$ and $\angle HPA=\angle QSM=90$ ie $\triangle QSM\sim \triangle AHP$ so $QS/QM=AP/AH$ since $AP=2*SP$ and $AH=2*OM$ ie $AP/AH=SP/OM$ but now $\angle QSP=\angle OMQ=90$ and $OM/SP=QM/QS$ so $\triangle OMQ\sim \triangle SPQ$ therefore $\angle SQP=\angle OQM=\angle OQB$ but since $AQ=QP$ ie $\angle AQP=2*\angle SQP$ so $\angle AQP=2*\angle OQB$

First, let use get the wording out of the way. Let $DEF$ be the orthic triangle. It is saying that $P$ is the intersection of $\odot AHEF$ with $BHC$. Now, we cant solve this straight away from here, we will need a few other stuff. Indeed, note that if $M$ is the midpoint of $BC$ that $EF$ is the polar of $M$, so it follows $\angle MFP = \angle FAP = \angle PBM$ (given) so the $P \in \odot MBF$ and do symmetry for other. Then, by radical axis, $PM \cap BF \cap BE = A$ so $P \in AM$. Now, we have enough so we ignore and focus on what we need to prove. We need $\angle QAM = \angle QOM \implies QAOM$ is cyclic. Let us prove $QA$ is tangent to $\odot ABC$. Since $Q$ is on the radical axis, it suffice to prove $QP$ is tangent to $\odot BHC$, and by symmetry this means if $AH \cap \odot BHC = A'$ then $QA'$ is tangent to $\odot BHC$. But, suppose the tangent at $P$ touches $BC$ at $Q'$. Then note $PBA'C$ is harmonic quad, so $Q'A'$ is tangent to $\odot BHC$ and by symmetry $Q'A = Q'P$ so $Q' = Q$ and we are done.

Let $O_1$ and $O_2$ be the circumcircles of $ABP$ and $ACP$, respectively. Notice that $O_1O_2$ is the perpendicular bisector of $AP$, so $O_1O_2 \cap BC = Q$. Let $M = AP \cap BC$. Let $Q'$ be on $BC$ such that $Q'A$ is tangent to the circumcircle of $ABC$. From $\angle PAB = \angle PBC$ and $\angle PAC = \angle PCB$, we know that the circumcircles of $ABP$ and $APC$ are both tangent to $BC$, and $AP$ is the radical axis, so that means $MB^2 = MC^2$, so $M$ is the midpoint of $BC$. We now show that $Q' = Q$. We have: \[\angle APB = 180^{\circ} - \angle BAP - \angle PBA = 180^{\circ} - \angle ABC \\ \angle APC = 180^{\circ} - \angle PAC - \angle PCA = 180^{\circ} - \angle BCA\] From Law of Sines, we have $O_1B = \frac{AB}{2\sin{ABC}}$ and $O_2C = \frac{AC}{2\sin{ACB}}$. Therefore, $\frac{QB}{QC} = \frac{AB \sin{ACB}}{AC \sin{ABC}} = \frac{AB^2}{AC^2}$. But since $Q'AB \sim Q'CA$, we have $\frac{Q'B}{Q'A} = \frac{AB}{AC}$ and $\frac{Q'A}{Q'C} = \frac{AB}{AC}$, and multiplying these two together yields $\frac{Q'B}{Q'C} = \frac{AB^2}{AC^2}$. Therefore, we have $Q = Q'$, so $QA$ is tangent to the circumcircle of $ABC$. But that means $OA \perp QA$ and $OM \perp BC$, so $QAOM$ is cyclic, so we know that $QO$ passes through the circumcenter of $AQM$. But $QO_1 \perp AP$, so $QO$ is the isogonal conjugate of $QO_1$ with respect to $AQM$, so $\angle AQO_1 = \frac{1}{2}\angle AQP = \angle OQB$, and multiplying that equation by $2$ gives us the desired conclusion. $\blacksquare$

The following solution is merely to prove the statements from the first given solution: Let $O_b, O_c$ be the circumcenters of triangles $ABP, APC$ and $AD$ altitude of $\triangle ABC$. We have $\angle BAO_b=\angle ABO_b=\angle BAD=\angle OAC\ (\ 1\ )$, $\angle CAO_b=\angle ACO_c=\angle DAC=\angle BAO\ (\ 2\ )$, from the above easily $\angle O_bAO=\angle DAO_c$, so $AO$ is angle bisector of $\angle O_bAO_c$; $O_bO_c$ being perpendicular bisector of $AP$, $Q, O_b, O_c$ are collinear, hence $\frac{CK}{BK}=\frac{CO_c}{BO_b}=\frac{AO_c}{AO_b}$, so $AK$ is external angle bisector of $\angle O_bAO_c$, i.e. $AQ\bot OA$, consequently $AOMQ$ is cyclic, of diameter $OQ$ and the altitude $QO_b$ of $\triangle AMQ$ is its isogonal w.r.t. $\angle MQA$, done. Best regards, sunken rock

Denote by $M$ the intersection of $AP$ and $BC$. We have $MB^2=MP*MA=MC^2$ so $M$ is midpoint of $BC$. Now let $P'$ be the reflection of $P$ in $BC$. $\angle BP'C=\angle BPC=180-\angle BAC$ so $P'$ is on circumcircle of triangle $ABC$. $\angle BMP'=\angle BMA$ so quadrilateral $ABP'C$ is harmonic hence tangents at $A$ and $P'$ (on circumcircle of $ABC$) meet $BC$ at one point,let's call it $Q'$. $Q'P=Q'P'=Q'A$ so $Q'=Q$. Quadrilateral $QAOM$ is cyclic so $90-\frac{1}{2}\angle AQP=\angle QAM=\angle QOM=90-\angle OQM$ so $2\angle QOM=\angle AQP$.

Seems too easy,so it might be wrong.It is easy to see that $<BPC=<180-<BAC$,now let $O'$ be the reflection of $O$ wrt $BC$.Now,we have to prove $<OQO'=PQA$ and since we have $QA=QP$ and $QO=QO'$,so it is enough to prove that triangles QPO' and QOA are congruent,but since we have QP=QA and QO=QO',it is enough to prove that $O'P = AO$,but this is true since $O'$ is the circumcentre of $BPC$ and $O'P=O'B=O'C=OB=OC=AO$,so we are finished.

junioragd wrote: Seems too easy,so it might be wrong.It is easy to see that <BPC=<180-<BAC,now let O' be the reflection of O wrt BC.Now,we have to prove <OQO'=PQA and since we have QA=QP and QO=QO',so it is enough to prove that triangles QPO' and QOA are congruent,but since we have QP=QA and QO=QO',it is enough to prove that O'P = AO,but this is true since O' is the circumcentre of BPC and O'P=O'B=O'C=OB=OC=AO,so we are finished. Yes it isn't very difficult, though it is very tempting to overkill it... Looking back at the problem, it is very natural to reflect $P$ about $BC$ to make $P'$, radical axis gives $OQ \perp AP'$, $\angle AQP = 2\angle AP'P$ so we need $\angle AP'P = \angle OQB$ which is obvious as $AP' \perp OQ, P'P \perp QB$.

Let line $AP$ meet $BC$ at $M.$ Then since $\angle MAB = \angle PBM$, we deduce that $\triangle MAB \sim \triangle MBP.$ By writing the length ratios, it follows that \[MB^2 = MP \cdot MA.\] Similarly, we find that $\triangle MAC \sim \triangle MCP$, and $MC^2 = MP \cdot MA.$ Hence, $MB = MC$, so $M$ is the midpoint of $\overline{BC}.$ In addition, note that \[\angle BPC = 180^{\circ} - \angle PBC - \angle PCB = 180^{\circ} - \angle PAB - \angle PAC - 180^{\circ} - A.\] It follows that the reflection $P'$ of $P$ about $BC$ lies on $\odot (ABC).$ Now, notice from $\triangle MAB \sim \triangle MBP$, we have $PB / AB = PM / BM.$ Similarly, we obtain $PC / AC = PM / BM.$ Therefore, \[1 = \frac{PB}{AB} : \frac{PC}{AC} = \frac{P'B}{AB} : \frac{P'C}{AC}\] which implies that the quadrilateral $BACP'$ is harmonic. Then let the tangents to $\odot (ABC)$ at $A$ and $P'$ meet at $Q'.$ Note that since $BACP'$ is harmonic, we have $Q' \in BC.$ Combining equal tangents with the fact that $BC$ is the perpendicular bisector of $\overline{AA'}$ and $\overline{PP'}$, it follows that \[Q'A = Q'P' = Q'P.\] Hence, $Q' \equiv Q.$ It follows that $QA \perp OA$, so the points $Q, A, O, M$ are inscribed in a circle of diameter $\overline{QO}.$ Then $\angle QAP = \angle QOM$, which implies that $90^{\circ} - \frac{1}{2}\angle AQP = 90^{\circ} - \angle OQM$, whence the desired result follows immediately. $\square$

This question must solve by the famous lemma. First we must know what is the point "$P$" this point is absolutely important.This point is on the median of $BC$. And we should know when we draw the circumcircle of the triangle $ABP$.that circumcircle tagents to $BC$.

Huyền Vũ wrote: Solution: (O2),(O1) are the excircles of triangle APB and APC Because of ∠PAB=∠PBC and ∠PAC=∠PCB so BC is the common tangent of (O1) and (O2). AP is the radical axis of (O1) and (O2) so AP intersecs BC at M where M is the midpoint of BC. So ∠OMQ=90 (1) Triangle QAB and QCA are similar so QA^2=QB*QC. Thus, QA is the tangent of the circle (O) or ∠OAQ=90 (2) From (1) and (2) O,A,Q,M are concyclic So ∠OMA=∠OQA But we also have ∠OMA=∠MQO1 (Q,O1,O2 are collinear) So ∠MQO1=∠AQO. So ∠MQO=∠AQO1. Final ∠AQP=2∠AQO1=2∠OQB. $Very Good$

Define $M$ as $\overline{AP}\cap\overline{BC}$, $N$ as the midpoint of $\overline{AP}$, and $\omega$ as the circle with diameter $\overline{BC}$. Clearly $\overline{QN}$ is the perpendicular bisector of $\overline{AP}$. Lemma: $\overline{AQ}$ is the tangent to $(ABC)$ at $A$ Proof: From $\triangle MPB\sim\triangle MBA$ and $\triangle MPC\sim\triangle MCA$, we can immediately obtain that $$MB^2=MP\cdot MA=MC^2,$$or that $M$ is the midpoint of $\overline{BC}$, and thus the center of $\omega$. Furthermore, this implies that $A$ is the image of inversion of $P$ wrt $\omega$, and thus $\overline{QN}$ is the radical axis of $\omega$ and the circle with radius $0$ at $A$. Now, since $\overline{BC}$ is the radical axis of $\omega$ and $(ABC)$, we must have that $Q$ is the radical center of $A$, $(ABC)$, and $\omega$, and so $Q$ lies on radical axis of $A$ and $(ABC)$, which is the desired tangent line. Now, to conclude, we note that $\overline{OM}\perp\overline{BC}$ and $\overline{OA}\perp\overline{AQ}$, so $A$, $Q$, $M$, and $O$ are concyclic, which implies that $$\angle AOQ=\angle AMQ=\angle NMQ.$$Combining this with $$\angle OAQ=\angle MNA=90^\circ$$yields that $$\frac{1}{2}\angle AQP=\angle NQA=\angle OQA-\angle OQN=\angle MQN-\angle OQN=\angle OQM=\angle OQB,$$and so $\angle AQP=2\angle OQB$. Q.E.D.

generization: conclusion is still true if P is any point on circle (BCH), H is orthocenter of ABC.

This solution is similar to what epitomy01 did, but perhaps shorter. Let $M$ be the intersection point of lines $AP$ and $BC$, it can be easily verified that $\bigtriangleup BPM$ and $\bigtriangleup ABM$ are similar, so $BM^2 = AM.PM$, in the same way we can get $CM^2 = AM.PM$, so $BM=CM$, i.e. $M$ is the midpoint of $BC$. Define $O_1$ and $O_2$ as the circumcenters of $\bigtriangleup ACP$ and $\bigtriangleup ABP$ respectively. From the fact that $Q, O_1$ and $O_2$ all lie on the perpendicular bisector of $AP$, then $Q, O_1$ and $O_2$ are collinear. Also note that $BO_2$ and $CO_1$ are perpendicular to $QC$ (by the fact that $\angle PBC = \angle PAB$ and $\angle PCB = \angle PAC$). Next, we have $\frac{O_{2}A}{O_{1}A} = \frac{O_{2}B}{O_{1}C} = \frac{QO_2}{QO_1} \Rightarrow QA$ is the external angle bisector of $\angle O_{2}AO_1$. Furthermore, we have $\angle BAO_2 = \angle ABO_2 = 90^\circ - B$ and in the same way, $\angle CAO_{2} = 90^\circ - C$. However, we know that $\angle BAO = 90^\circ - C = \angle CAO_1 \Rightarrow \angle OAO_1 = \angle BAC$, in the same way we get $\angle OAO_2 = \angle BAC$, so $AO$ is the internal angle bisector of $\angle O_{2}AO_1$, hence $\angle QAO = 90^\circ$, i.e. $QA$ is tangent to $(ABC)$. Now, observe that $AQMO$ is cyclic since $\angle QAO = \angle OMQ = 90^\circ$. This implies that $\angle OQM = 90^\circ - \angle QOM = 90^\circ - \angle QAM = \angle AQO_2$. But since $AQ=QP$, then $\angle PQO_2 = \angle AQO_2 = \angle OQM \Rightarrow \angle QAP = 2\angle OQB$, as desired.

let $\angle AQP=2x, \angle OQB=y$ and $M$ is the midpoint of $BC$ it's well-known that $P$ is the $A-HM$ point claim:$P$ is on the A-appolonain circle $\omega$ proof: Do $\sqrt{bc}$ inversion then $(\omega)^*$ is the perpindacular bisector of $BC$ but $P^*$ is the intersection of tangents from $B,C$ to $(ABC)$ $\blacksquare$ thus $AQ$ is tangent to $(ABC)$ $\angle OMQ= \angle OAQ=90 \implies AOMQ$ is cyclic then $90-x=\angle QAP=\angle QOM=90-y$ and we win

Since no one posted this elegant solution... Let $P'$ be the reflection of $P$ over $\overline{BC}$. Let $R$ be the intersection of the tangents to $(ABC)$ at $B$ and $C$. Let $M$ be the intersection of lines $AP$ and $BC$. Since $P$ is the $A$-Humpty point of $\triangle ABC$, $M$ is the midpoint of $\overline{BC}$ and $\angle QMO = 90^{\circ}$. Thus $\angle AQP = 2\angle OQB$ if and only if $\angle QAP = \angle QAM = \angle QOM$. Thus we wish to prove $QAOM$ is cyclic $\implies\angle OAQ = 90^{\circ}\implies\overline{QA}$ is tangent to $(ABC)$. Note that since $\angle BP'C=\angle BPC=180^{\circ}-\angle BAC$, $P'$ lies on $(ABC)$. Let $Q'$ be the intersection between the tangents to $(ABC)$ at $A$ and $P'$. Proving $Q'=Q$ thus finishes. Note that since $P$ lies on the $A$-Apollonius circle of $\triangle ABC$, $\tfrac{AB}{AC}=\tfrac{PB}{PC}=\tfrac{P'B}{P'C}$. Thus $ABP'C$ is harmonic, and $A$, $P'$, $R$ are collinear. Thus by La Hire's theorem, since $R$ lies on the polar of $Q'$, $Q'$ must lie on the polar of $R$ with respect to $(ABC)\implies Q'$ lies on line $BC$. Further, note that $Q'A=Q'P'=Q'P$, thus $Q'$ also lies on the perpendicular bisector of $\overline{AP}$. Thus $Q'=Q$, as desired. $\blacksquare$

Claim: $QA$ is tangent to $\odot (ABC)$. Proof: It is well known that $AP \cap BC=M$ the midpoint of $BC$. Consider an inversion with center $M$ and radius $MB$. This swaps $A$ and $P$, so $QN$ is the radical axis of the degenerate circle at $A$ and the circle with diameter $BC$. $$QB \times QC=QA^2$$and the result follows. Then since $\angle OMQ=\angle QAO=\angle QAN=\frac{\pi}{2}$ $$\angle AQN=\angle MAO=\angle OQM$$ and we are done.

Solution: We can easily see that $P$ is the $A$-humpty point of $\triangle{ABC}$ so $P$ lies on the $A$-median. Let $AP \cap BM = D$. By the problem statement, we need $90 - \frac{\angle{BQA}}{2} = 90 - \angle{OQB} \implies \angle{QAD} = \angle{DOQ} \implies ODAQ$ would be cyclic, so $\angle{OAD} = 90$, or $AQ$ is tangent to the circumcircle of $\triangle{ABC}$. Thus, this prompts us to use Phantom Points, i.e. take some point $R$ s.t $RA$ is tangent to the circumcircle of $\triangle{ABC}$ and try to prove that $RA = RP$. From the properties of the Humpty point, we have $\frac{AB}{AC} = \frac{PB}{PC}$ and we also know that $\frac{AB}{AC} = \frac{BR^2}{CR^2} \implies \frac{PB}{PC} = \frac{BR^2}{CR^2}$. Now we will show that this leads to the fact that $RP$ is tangent to the circumcircle of $\triangle{PBC}$. As a matter of fact, if we let the tangent from $P$ to the circumcircle of $\triangle{PBC}$ intersect $BC$ at some point $S$, then we have $\frac{PB}{PC} = \frac{BS^2}{CS^2} \implies \frac{BR^2}{CR^2} = \frac{BS^2}{CS^2} \implies R = S$. Finally, by POP, we have $RA^2 = RB*RC = RP^2 \implies RA = RP$, as claimed. $\square$

$P$ is $A$-humpty point of $\triangle{ABC}$ so $P$ lies on circle $A-Appolonius$ with ratio $\frac{AB}{AC}$ (familiar result). We know that the center of $A-Appolonius$ is the intersection of tangent at $A$ of $(ABC)$ and $BC$, so that's $Q$. Let $M,N$ be midpoint of $BC, AP$. We have $\overline{A,N,P,M}$ We need $\measuredangle OQB =\frac{\measuredangle AQP}{2}=\measuredangle AQN\Leftrightarrow \bigtriangleup AQN \sim \bigtriangleup OQM $ But $\measuredangle OQB =\frac{\measuredangle AQP}{2}=\measuredangle AQN\Leftrightarrow \measuredangle QAM =\measuredangle QOM \Leftrightarrow QAOM$ is cyclic, which is exactly true. (Q.E.D)

Fang-jh wrote: Let $ABC$ be an acute scalene triangle with $O$ as its circumcenter. Point $P$ lies inside triangle $ABC$ with $\angle PAB = \angle PBC$ and $\angle PAC = \angle PCB$. Point $Q$ lies on line $BC$ with $QA = QP$. Prove that $\angle AQP = 2\angle OQB$. By the angle condition we can deduce that $P$ is the A-humpty point, let $M$ the midpoint of $BC$, let $A'$ the 2nd point in $(O)$ that lies on the A-symedian, let $D$ the midpoint of $AP$ and $D'$ the midpoint of $PA'$. Its known that $P$ and $A'$ are symetric w.r.t. $BC$ hence $D'$ lies on $BC$ and $PA'$ is perpendicular to $BC$ and $QA=QP=QA'$ meaning that $Q$ is the intersection of the tangents from $A$ and $A'$ to $(O)$, hence $AOMA'Q$ is cyclic, now let $QD$ meet $PD'$ at $L$, by angle chasing: $$\angle QLA'=\angle AMQ=\angle QMA' \implies O,A,L,M,A',Q \; \text{concyclic}$$Now the problem is equivalent to show that $\angle AQD=\angle OQB$ but now since $AM \parallel LO$ we have that arcs $AL$ and $OM$ are equal in $(OALMA'Q)$ hence $\angle AQD=\angle OQB$ holds, thus we are done

Let $M$ be the midpoint of $BC.$ Clearly $\odot (ABP),\odot (ACP)$ tangent to $BC,$ so $M\in AP$ and $Q$ is the exsimilicenter of these circles. Now from $|QA|^2=|QB|\cdot |QC|$ we get $Q\in \odot (AOM),$ and hence $$\angle AQP=180^\circ-2\angle QAP=180^\circ -2\angle QOM=2\angle OQB \text{ } \blacksquare$$

Headsolves for the win . Let $M$ be the midpoint of $BC$, ray $AM$ meet $(ABC)$ again at $R$, $T = BB \cap CC$, $K = AT \cap BC$, $N = \overline{AKT} \cap OQ$, and the projection of $A$ onto $BC$ be $X$. The angle conditions imply $P$ is the $A$-Humpty point, so $P$ lies on $\overline{AMR}$. Now, a well-known lemma regarding the $A$-Apollonian Circle wrt $BC$ yields $Q = AA \cap BC$. Thus, $N$ is actually the $A$-Dumpty point, which means $BNOCT$ is cyclic with diameter $OT$. Because $AK$ and $AR$ are isogonal in $\angle BAC$, angle chasing yields $ABK \sim ARC$. Thus, $$\angle OQB = 90^{\circ} - QOM = 90^{\circ} - \angle NOT = \angle NTO = \angle KAX$$$$= 90^{\circ} - \angle AKB = 90^{\circ} - \angle ACR = 90^{\circ} - \angle QAR = \frac{180^{\circ} - 2 \angle QAP}{2}$$$$= \frac{180^{\circ} - \angle QAP - \angle QPA}{2} = \frac{\angle AQP}{2}$$as desired. $\blacksquare$ In Retrospect: One can just notice $AOMQ$ is cyclic, implying $$2 \measuredangle OQB = 2 \measuredangle OAM = 2(90^{\circ} - \measuredangle MAQ) = 2 \measuredangle QAM = \measuredangle QAP + \measuredangle APQ = \measuredangle AQP$$which is much simpler.

Hah, lengths. Let $M$ be the midpoint of $\overline{BC}$; it's known/not too hard to prove $\triangle ABM\sim\triangle BPM$ and $\triangle ACM\sim\triangle CPM$. This means \[ \frac{PB}{PC} = \frac{PB/PM}{PC/PM} = \frac{AB/BM}{AC/CM} = \frac{AB}{AC}. \]Now define $Q_0:= AA\cap BC$. We claim that $Q_0A = Q_0P$. To prove this, define the function $f$ via \[ f(X) = PX^2 - \operatorname{pow}_{\odot(ABC)}(X); \]remark that $f$ is an affine map. Observe $f(B) = PB^2$, $f(C) = PC^2$, and \[ \frac{BQ_0}{Q_0C} = \frac{AB^2}{AC^2} = \frac{f(B)}{f(C)}. \]It follows that $f(Q_0) = 0$, i.e. $Q_0A = Q_0P$. Thus $Q_0\equiv Q$, ergo $QAOM$ is cyclic. The desired equality follows by an angle chase.

From the angle conditions, it is clear that $P$ is the $A$-Humpty point. Since $A$ and $P$ both lie on the $A$-Apollonius circle, $Q$ must be the center. It is well known this circle is orthogonal to $(ABC)$, so $\angle QAO = 90$. Denote $M = AP \cap BC$ as the midpoint of $BC$. We then have $MOAQ$ cyclic, and hence \[\angle MAQ = \angle MOQ \implies \angle PAQ = 90 - \angle OQB \implies \angle APQ = 2 \angle OQB.~\blacksquare\]

Lets name the circumcircle of $ABC$ as $\omega$. If we denote the intersection of tangents to the $\omega$ at $B$ and $C$ as $S$ , the midpoint of $BC$ as $M$, orthocenter as $H$, and the intersection of $\omega$ and $AS$ as $L$. It's well know that $P$ is the A-Humpty point of triangle $ABC$ so $A$, $P$, $M$ are collinear and $HP$ will be perpendicular to $AM$. Lets take $Q$ as the intersection of the tangent at $A$ wrt $\omega$ with $BC$. Then $Q$ lies on $BC$ which implies $S$ lies on the polar of $Q$ wrt $\omega$ which implies that $QL$ is tangent to $\omega$. $AL$ is isogonal to $AM$ so that implies that $\angle PBC = \angle LBC$ and $\angle PCB = \angle LCB$ which leads to $L$ being the reflection of $P$ wrt $BC$ which in turn implies that $QP = QL = QA$. Thus if we take midpoint of $AP$ as $R$, $\angle ARQ = \angle OAQ = 90$ which makes $\angle OAM = \angle AQR$. $\angle OMQ = \angle OAQ = 90$ so $AOMQ$ is cyclic implying $\angle OAM = \angle OQM = \angle AQR = \frac{1}{2}\angle AQP$ finishing the proof.

Let $S$ be intersection of $A$ - symmedian of $\triangle ABC$ with $(O)$. From the hypothesis, we have $P$ is $A$ - Humpty point of $\triangle ABC$. So $Q$ must be the intersection of tangent at $A$ of $(O)$ with $BC$. Then $OQ \perp AS$. We have the familiar result that $PS \perp BC$. Hence $\angle{AQP} = 2\angle{ASP} = \angle{BQO}$

If we add circumcenter $O'$ of $BHC$, then we have $O'P=O'B=OB$(because P lies on circumcircle of $BHC$ and $O'$ is reflected of $O$ with respect to $BC$). so we have $QO=QO', QA=QP, AO=O'P$ so $OAQ$ and $O'PQ$ are similar. So $QAP$ and $QOO'$ are similar too(why?) So we have $OQO'=2OQB=AOP$ and were done!!

Note that $P$ is the $A$-Humpty point. Take $\sqrt{bc}$ inversion, then $P$ goes to $T=BB \cap CC$. The image of $Q$ lies on the circle centred at $T$ passing through $A$, and on $(ABC)$, so it is the reflection of $A$ across the perpendicular bisector of $BC$. Call this point $A'$. Inverting back, we get that $Q=AA \cap BC$. Let $M$ be the midpoint of $BC$, then since $\angle QAO = 90^\circ = \angle QMO$, $(AOMQ)$ is concyclic. Let $H$ be the orthocentre, then $$\frac{1}{2} \angle AQP = \frac{1}{2} \angle ATA' = \angle ATO = \angle TAH = \angle OAM = \angle OQM = \angle OQB$$so we are done. $\square$

$P$ is $A-$humpty point. Perform $\sqrt{bc}$ inversion and reflect over the angle bisector of $\measuredangle CAB$. New Problem Statement: $ABC$ is a triangle $(AB<AC)$ where $T$ is the intersection of tangents to $(ABC)$ at $B,C$ and $D\in (ABC)$ with $AD\parallel BC$. $A'$ is the reflection of $A$ over $BC$. Prove that $\measuredangle ATD=2\measuredangle B-2\measuredangle C-2\measuredangle AA'D$. Let $M$ be the midpoint of $BC$ and $S$ be the intersection of $A-$symmedian with $(ABC)$. Since $\measuredangle DAA'=90$ and $MD=MA=MA'$ we see that $D,M,A'$ are collinear. Also \[(BC_{\infty},M;B,C)=-1=(A,S;B,C)=(DA,DS;DB,DC)=(BC_{\infty},DS\cap BC;B,C)\]implies $D,M,S$ are collinear. \[\frac{\measuredangle ATD}{2}+\measuredangle AA'D=\measuredangle ATM+\measuredangle TMA'=\measuredangle ASD=\measuredangle B-\measuredangle C\]As desired.$\blacksquare$