My solution: Let $ T $ be the center of $ \mathcal{W} $ . From $ \angle HDC=\angle HAB=\angle HCD \Longrightarrow D $ is the reflection of $ C $ in $ AH $ . Since $ \angle APD=\angle ACB+\angle CDP=\angle ACB+90^{\circ}-\angle CBA=90^{\circ}-\angle DAB $ , so $ AB $ pass through the center $ Q $ of $ \odot (ADP) \Longrightarrow AB $ is the diameter of $ \odot (ADP) $ , hence from $ \angle DTB=2\angle DAB=\angle DQB $ we get $ T $ lie on $ \odot (BDQ) $ . Q.E.D

Dear Mathlinkers, the main idea of this problem is based on the Morley's circle... See: http://jl.ayme.pagesperso-orange.fr/Docs/Les%20cercles%20de%20Morley,%20Euler,%20Mannheim%20et%20Miquel.pdf Sincerely Jean-Louis

my solution = let $O$ be circumcentre of triangle $ABD$ than clearly $AD$ is radical axis of of circle of $ABD,ADP$ and hence $AD$ is perepndicular to $OQ$ and thus $\angle OQD=\angle APD$ now angle chasing gives $\angle DAP=180-2C$ and $\angle HBA=\angle ADH=90-A$ and hence $\angle APD=\angle OQD = C-B+90$ now $\angle BOD=2\angle BAD=2(A+C-180)$ and hence $\angle OBD=90-\angle BOD=90+B-C$ and thus $\angle OBD+\angle OQD=(90+B-C) +(90+C-B) = 180$ and hence $BDQO$ is cyclic quad. we are done

Let $O'$ be the circumcenter of $\triangle ADP$. Note that $O'Q$ is the perpendicular bisector of $AD$ and let it meet $BC$ at $T$. Now by power of a point, it shall suffice to showing $TD.TB=TO'.TQ$ Now just trig bash!!!

Dear Mathlinkers, I use http://www.artofproblemsolving.com/community/c6h1142448_a_circumcenter_on_bc as a lemma and conclude with the Morley cicle mantioned above. Sincerely Jean-Louis

Let $BH\cap CA=E$. $\angle DAB=\angle DHB$(A,H,B,D concyclic) $=\angle PHE=90^{\circ}-\angle APD.$ Also, $Q$ being circumcentre of $\triangle ADP$, we have $\angle DAQ=90^{\circ}-\angle APD=\angle DAB.$ So $Q$ lies on $AB$, giving $\angle BQD=2\angle DAB=\angle DOB$, where $O$ is the circumcentre of $\triangle ADB$. Hence proved.

Three years ago, I cartesian bashed this problem. My re-solve is the same as above

Dear Mathlinkers, http://jl.ayme.pagesperso-orange.fr/Docs/Orthique%20encyclopedie%201.pdf p. 16... Sincerely Jean-Louis

WLOG, Assume $D$ lies outside $\Delta ABC$ (Though this also works for $D$ inside $\Delta ABC$). Let the perpendicular to $AC$ at $P$ intersect $AB$ at $A'$, then, $\angle ADP$ $=$ $\angle ABH$ $=$ $\angle AA'P$ $\implies$ $D$ $\in$ $\odot (AA'P)$. Hence, $AA'$ is the diameter of $\odot (ADP)$ $\implies$ $Q$ $\in$ $AB$. Let $O'$ be the center of $\odot (ABH)$ $\implies$ $\angle DO'B$ $=$ $\angle DAB$ $=$ $\angle DQB$ $\implies$ $O'$ $\in$ $\odot (BDQ)$

Claim: $\angle DHF=\angle B$. Proof: $\angle DHF=180-\angle DHA=180-\angle DBA=\angle B$. $\square$ Claim: $Q\in AB$. Proof: Now, \begin{align*} \angle DAB&=\angle DHB=\angle DHF-\angle BHF=\angle B-(90-\angle HBC)=\angle B-\angle C \\ \angle DAQ&= \tfrac12(180-\angle DQA) = 90-\angle DPA = 90-(180-\angle AHP-\angle PAH) \\ &= 90-(180-\angle DHF - (90-\angle C)) \\ &= 90-(180-\angle B-(90-\angle C)) \\ &= \angle B-\angle C. \end{align*}Hence $\angle DAB=\angle DAQ$, so $Q\in AB$. $\square$ We want to show $O'BDQ$ is cyclic, where $O'$ is the circumcenter of $(ABH)$. We have \begin{align*} \angle DO'B&=2\angle DAB \\ \angle DQB&=180-\angle DQA=2\angle DAQ=2\angle DAB. \end{align*}The second equality follows since $QA=QD$ and $Q\in AB$. We are done.

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Let $F=AB\cap (PAD)$. Notice $\angle DHF=\angle DAF=\angle DAB=\angle DHB$, which means $BH\parallel FP$. Since $BH\perp AC$ therefore $FP\perp AC$. Meaning $Q$ lies on $AF\equiv AB$. Let $E$ be the circumcenter of $(ABH)$. Now $\angle BED=2\angle BAD=\angle BQD$.

We ignore configuration issues. I think this is for $\angle C>\angle B$ and everything acute. $\angle CDP=\angle HAB=90^\circ-\angle B$, so $\angle CPD=90^\circ+\angle B-\angle C$. Thus $\angle AQD=180^\circ+2\angle B-2\angle C$, and $\angle DAQ=\angle C-\angle B$. On the other hand, $\angle DAB=\angle DHB=180^\circ-(90^\circ-\angle C)-\angle BDH=90^\circ+\angle C-180^\circ+(90^\circ-\angle B)=\angle C-\angle B$, so $Q$ lies on $\overline{AB}$. Thus $\angle BQD=2\angle C-2\angle B$. It thus suffices to show that $\angle BAD=\angle C-\angle B$, or $\angle BDA=180^\circ-\angle C$, but this follows since $\angle BDA=\angle BHA$. $\blacksquare$