Let $\omega$ and $(ACB)$ meet at $X$; clearly $ADX$ collinear. Let $CX$ meet $\omega$ at $Q'$; we aim to show $PQ'$ is tangent. Call $M = \overline{AD} \cap \overline{CH}$. Let the line through $P$ parallel to $AD$ meet $AB$ at $T$. Note that $\triangle ACB \sim \triangle DQ'P$ through a spiral similarity at $B$. Since $PD/DB = TA/AB$, this spiral similarity takes $T$ to $P$. Projecting $(AM;D\infty)=-1$ onto $AB$ gives $(TH;AB)=-1$ hence $TC$ is tangent to $(ACB)$, and done.

[asy][asy] unitsize(3cm); pointfontpen=fontsize(10); pointpen=black; pathpen=rgb(0.4,0.6,0.8); pair B=dir(0), A=-B, C=dir(110), H=foot(C,A,B), N=WP(D(C--H),0.85), D=2*N-A, P=extension(B,D,C,H), Q=tangent(P,(B+D)/2,length(B-D)/2,2), K=extension(A,D,C,Q), M=(B+D)/2; D(CR((0,0),1,0,180),heavygreen); D(arc(M,length(K-M),degrees(B-M),degrees(D-M),CCW),red); D(A--B--C--cycle,linewidth(1)+pathpen); DPA(D--B--Q--cycle^^Q--P--D^^H--C); D(C--K--A,dashed+pathpen); /*Angle marks and pathticks*/ markscalefactor=0.01; DPA(rightanglemark(B,C,A)^^rightanglemark(B,H,C)^^rightanglemark(B,Q,D)); /* Dot and label points */ D("A",A,W); D("B",B,E); D("C",C,dir(C)); D("D",D,SSE); D("H",H); D("K",K,dir(K)); D("P",P,dir(P)); D("Q",Q,dir(Q)); [/asy][/asy] Let $K=AD\cap CQ$. We claim that $\triangle ABC\sim\triangle DBQ$. Indeed, $\angle BCA=\angle BQD=90^{\circ}$ and since $\triangle PDQ\sim\triangle PQB$ and $CH$ bisects $\overline{AD}$, $$\frac{DQ^2}{BQ^2}=\frac{PQ^2}{PB^2}=\frac{PD}{PB}=\frac{AH}{BH}=\tan^2B=\frac{AC^2}{BC^2},$$and our claim holds. Thus $B$ is the center of the spiral similarity $\overline{AD}\to\overline{CQ}$, so $\angle KDB=\angle KQB$, and $K$ lies on $\omega$.

Let $S$ be the foot from $Q$ to $BD$, and let $E$ be the foot from $D$ to $AB$. Note that since $AH$ bisects $AD$, $H$ is the midpoint of $AE$. We have that $\frac{DS}{BS}=\frac{DP}{BP}=\frac{EH}{BH}=\frac{AH}{BH}$. Since $AC\perp BC$ and $DQ\perp BQ$, we have that $BQD$ and $BCA$ are similar. The spiral similarity sending $QD$ to $CA$ is centered at $B$, so $AD\cap CQ$ lies on the circumcircle of $BDQ$, as desired.

Alternative Finish: We note as above that $DQ/QB = AC/CB$. As $\angle DQB = \angle ACB = 90$, $B$ is the center of spiral similarity mapping $DQ$ to $AC$. Then by a well known lemma, it follows that the intersection of $CQ$ and $AD$ must lie on the circumcircles of $ACB$ and $DQB$. In particular, this $(DBQ) = \omega$ so we are done.

Denote $T=AD \cap w$ and $Q'=CT \cap w$. Let $X= AB \cap w$ and let $Y= CB \cap Q'D$. Simple angle chasing yields to $CPDY$, $CTBA$, cylic. Furthermore $\angle Q'BY= \angle DBX$ and $\angle PYQ'= \angle CDX$. Also $\angle ACH= \angle HCX= \angle CXD$. Easily $\triangle YQ'B \sim \triangle DXB$ and $\triangle BYP \sim \triangle BDC$ $\Longrightarrow \frac{YQ'}{DX} = \frac{YB}{DB} = \frac{PY}{CD}$ By SAS we get $\triangle CDX \sim \triangle PYQ' \Longrightarrow \angle CXD= \angle PQ'Y$. $\Longrightarrow \angle PQ'Y= \angle CXD= \angle XCH= \angle ACH= \angle CBA = \angle CTA$. Thus, $PQ'$ is tangent to $w$ and $Q=Q'$. Done!

Let $M$ be the midpoint of $AD$, $F$ is the reflection of $C$ across $M$. $AD,BC,ME$ meets $\omega$ at $X,E,Q'$ respectively, tangent to $\omega$ at $Q'$ meets $BD$ at $P'$. Note $CAFD$ is a parallelogram $\implies F,D,E$ collinear. $\angle CEF=90^{\circ}\implies MC=ME=MF$. $A,C,X,B$ are cyclic with diameter $AB$, so $\angle CXA=\angle CBA=90^{\circ}-\angle MCB=90^{\circ}-\angle MEC=\angle MED=\angle Q'XD=\angle Q'XA\implies C,Q',X$ collinear. Pascal's on hexagon $Q'Q'EBDX\implies C,P',M$ collinear thus $P'\equiv P\implies Q'\equiv Q$. Hence $CQ,AD$ meet at a point $X$ on $\omega$.

Let $Q' \in \omega$ such that $B$ is the spiral center sending segment $DQ'$ to $AC$. Then $F\equiv AD\cap PQ' \in \omega, \odot(ABC)$. Letting $P'$ be the image of $P$ under the aforementioned spiral similarity, it follows that segments $AD$ and $P'P$ are homothetic at $B$, so defining $S\equiv AD\cap CH$ gives $$-1=(A,D;S,P_{\infty})\stackrel{P}{=}(A,B;H,P')$$so $P'$ lies on the polar of $H$ with respect to $\odot(ABC)$. This gives the desired conclusion.

Let $E$ be the projection of $D$ on $BC$. From $\frac{AC^2}{AB^2}=\frac{HA}{HB}=\frac{HE}{HB}=\frac{PD}{PB}=\frac{QD^2}{QB^2}$ implies $\triangle DBQ\sim \triangle ABC$. Hence, $\measuredangle CBQ=\measuredangle DBA,\frac{DB}{BQ}=\frac{BA}{BC}$ so $\triangle DBA\sim \triangle QBC$ i.e $\measuredangle QCB=\measuredangle DAB.\blacksquare$

I think this problem is richer than I initially thought, especially in solution 3 here Solution 1: Let $AD\cap CH=X$, let $Y$ be a point on $AB$ such that $PY\parallel AD$. Observe that $$-1=(A,D;X, \infty)=(A, B;H,Y) $$So this implies that $CY$ is tangent to $(ACB)$, since $CH$ is the symmedian for $\triangle ACB$. Define $R=AD\cap (ACB)$, clearly $R$ lies on the circle with diameter $DB$, and redefine $Q=CR\cap (RDB)$. By spiral similarity lemma, $\triangle DQB\sim \triangle ACB$, so to prove $PQ$ is tangent to $(DQB)$, just need to prove $Y$ maps to $Q$ in this spiral similarity. But this is because $YP\parallel AD$, so $\frac{BA}{BY}=\frac{BD}{BP}$. So we are done. Solution 2: Like above, we prove $\triangle DQB\sim \triangle ACB$, Now, by Menelaus Theorem, if $AD\cap CH=X$, then $$\frac{DP}{PB}\cdot \frac{BH}{HA}\cdot \frac{AX}{XD}=1$$$$\Rightarrow \frac{AH}{HB}=\frac{PD}{PB}$$But then $\frac{PD}{PB}=\frac{AH}{AB}=(\frac{AC}{AB})^2=(\frac{DQ}{QB})^2$, so $PQ$ is tangent to $(DQB)$. Solution 3: Let $C'$ be reflection of $C$ about $AB$ and $G$ is reflection of $F'$ about $BD$. Observe $C'$ is image of $G$ under $\phi$, so $C', G, E$ colinear. Meanwhile, if $A'$ is such that $AH=HA'$, then the given condition means $A'$ lies on $(BF'D)$ because $DA'\perp AB$. So $\angle C'A'A=\angle CAB=\angle FDB=\angle FA'B$ thus $F', A', C'$ are colinear. Also observe $\angle GA'B=\angle GDB=\angle F'DB=\angle CAB=\angle CA'A$, so $C, A', G$ colinear. Now look at the complete quadrilateral $EC'GCA'F'$, and consider the pole of $F'G$, clearly it lies on perpendicular bisector of $BD$ since $F'$ and $G$ are symmetric about $BD$. Also, the pole of $F'G$ lies on polar of $F'G\cap AE$, which by Brokard's theorem its $CC'$ because $C=GA'\cap F'E$ and $C'=F'A'\cap GE$. So the pole of $F'G$ is $CC'\cap BD=P$, which means $PF'$ is tangent to $(DQ'B)$.

Let $T$ be the intersection of $\omega $ and $AB $. Then we have that $DT \parallel PH $. Since $PH $ bisects $AD $ $\implies CH $ bisects $AT $. $\implies AH=HT $ Hence $\widehat {HTC}=\widehat {HAC}=\widehat {HCB}$ Let $\omega $and $(ABC $ meet at $X $ and let $E $ be the reflection of $C $ upone $BC$. Let $Q'$ bethe second intersection of $ET $ and $\omega $. CLAIM $C.Q',X $ are collinear Proof... $ \widehat {CXB} =180^{\circ} -\widehat {CEB}=180^{\circ} -\widehat {HCB}=180^{\circ} -\widehat {HTC}=180^{\circ} -\widehat {HTE}=180^{\circ} -\widehat {Q'TB}= \widehat {Q'XB} $ Hence it implies $\implies $ $C,Q',X $ are collinear . $\implies $ $Q'=Q $ CLAIM- $EPQB $ are cyclic Proof...Since $PH \parallel DT$ and $TDQB $ are cyclic We have that $\widehat {EPB}=\widehat {TDB}=\widehat {TQB} $ $\implies $ $EPQB $ are cyclic CLAIM- $\widehat {QBP} =\widehat {HBE }$ Proof... since $PQBE $ are cyclic and $E$ is the reflection of $C $. $\widehat {QBP}=\widehat {QEP}=\widehat {HTC}=\widehat {HBC}=\widehat {HBE} $ Hence we have that $\widehat {PQE}=\widehat {PBE}=\widehat {QBT} $ $\implies $ $PQ $ is tangent to $\omega $.

Let $\{ F\}=AD\cap CH,\ \{E\}=BD\cap (ABC)$. We have $\widehat{AEB}=90^\circ$, so $E\in (BDQ)$, hence it's enough to prove $C-Q-E$ collinear. As $PQ$ is tangent to $(QDB)$, $\dfrac{PB}{PD}=\left ( \dfrac{QB}{QD} \right ) ^2$. By Menelaus in $\triangle{ADB}$ we get $\dfrac{PD}{PB}=\dfrac{HA}{HB}$. As $\triangle{CAB}$ is right angled, $\dfrac{AH}{BH}=\dfrac{\frac{AC^2}{AB}}{\frac{BC^2}{AB}}=\left ( \dfrac{AC}{BC} \right )^2$. Thus, $\dfrac{QD}{QB}=\dfrac{AC}{BC}$, so $\triangle{ACB}\sim \triangle{DQB}$, whence $\widehat{QBD}=\widehat{CBA}$. We have $$\widehat{CEA}=\widehat{CBA}=\widehat{QBD}=\widehat{QED}=\widehat{QEA}$$thereby $C-Q-E$ are collinear.

We basically want to show that $B$ is the Miquel point for the quadrilateral $ACQD$ since then it follows that $CQ,AD$ concur on $\omega$. Let $S$ be the point at which the tangent to $(ACB)$ at $C$ meets $AB$. We prove that $\frac{DP}{BP}=\frac{AS}{BS}$. Clearly, this will establish the aforementioned result. Let $D'$ be the foot of altitude from $D$ onto $AB$. Notice that in right angle triangle $CAB$ (right angled at $C$) the altitude $CH$ is a symmedian from $C$. Therefore, $(A,B;H,S)=-1$. It follows that $\frac{AS}{BS}=\frac{AH}{BH}=\frac{HD'}{BH}=\frac{DP}{BP}$ and we conclude. Comment. This problem featured in Iran TST if I know correctly.

Let $R$ be the intersection of the 2 semicircles. $AD$ intersects $(ABC)$ at $R$. Then $R$ is on $(DB)$. $Q'$ is the intersection of $CR$ and $(BD)$. So $BQ'D$ and $BCA$ are similar. So we get that $\frac{Q'D}{PD}= \frac{BD \cdot AC}{PD \cdot AB}= \frac{CA}{MA}$ where $M$ is the intersection of the tangent at $C$ and $AB$. Thus $PQ'$ is tangent to $(BD)$ and we are done

That was also German TSTST #6.

We can go ahead and just prove $\triangle BQD \sim \triangle BCA$ since that will give us $\triangle BQC \sim \triangle BDA$. To do this, we calculate $\frac{BQ}{QD} = \frac{PQ}{PD}=\sqrt{\frac{PB}{PD}} = \sqrt{\frac{BH}{AH}} = \frac{BC}{CA}$.

Again UK TST 2 Q6. Couldn't solve it in the exam though.

[asy][asy] unitsize(3cm); pointfontpen=fontsize(10); pointpen=pathpen=black; pair B=dir(0), A=-B, C=dir(120), H=foot(C,A,B), N=WP(C--H,0.85), D=2*N-A, P=extension(B,D,C,H), Q=tangent(P,(B+D)/2,length(B-D)/2,2), K=extension(A,D,C,Q), T=2/(C+1/C), M=(B+D)/2; DPA(CR((0,0),1,0,180)^^arc(M,length(K-M),degrees(B-M),degrees(D-M),CCW)^^A--B--C--cycle^^D--B--Q--cycle^^Q--P--D^^C--T--A); DPA(H--C--K--A ,dashed); /*Angle marks and pathticks*/ markscalefactor=0.01; DPA(rightanglemark(B,C,A)^^rightanglemark(B,H,C)^^rightanglemark(B,Q,D)); /* Dot and label points */ D("A",A,SW); D("B",B,SE); D("C",C,dir(C)); D("D",D,SSE); D("H",H); D("K",K,dir(K)); D("P",P,dir(P)); D("Q\: (Q')",Q,dir(Q)); D("T",T,W); [/asy][/asy] Let $T$ be the inverse of $H$ in $(ABC)$. Let $K=AD\cap\omega, Q'=CK\cap\omega$. It suffices to show that $PQ'$ is tangent to $\omega$. But $\frac{PD}{PB}=\frac{HA}{HB}=\frac{TA}{TB}$, so $B$ is the center of the spiral similarity $\triangle DBQ'\cup P\to\triangle ABC\cup T$, and the result follows.

We label our points as in jlammy's figure above. Notice that the statement is equivalent to $\triangle BQD \sim \triangle BCA$. Let $\phi$ denote the spiral similarity centered at $B$ sending $A$ to $D$. Let $Q'=\phi(C)$, $BD\cap CH=P$, and let the tangent to $\omega$ at $Q'$ meet $CH$ at $P'$. We will show that $P=P'$. I now give a synthetic sketch. Let $\odot ABC$ be the unit circle in $\mathbb C$. Let the reflection of $C$ across $AB$ be $C'$. Then, $c'=\frac{b^2}{c}$. Let $X$ be an arbitrary point on $CC'$. We have that $$x=c+\frac{b^2}{c}-2b^2\overline{x} \implies \overline{x}=\frac{c}{b^2}+\frac{1}{c}-\frac{x}{b^2}.$$Then let $D$ be the reflection of $A$ over $X$. So $d=2x+b$. Now we compute $\phi(z)=-\frac{x}{b}(z-b)+b.$ To compute $p'$, we compute $\phi(t)$. We have that $t=\frac{2cc'}{c+c'}=\frac{2b^2c}{b^2+c^2} \implies p'=\phi(t)=- \frac{x}{b}\left(\frac{2b^2c}{b^2+c^2}-b\right)+b=\frac{-2xbc+xb^2+xc^2+b^3+bc^2}{b^2+c^2}.$ Now using the intersection of lines formula, we get \begin{align*} p&=\frac{(\overline{b}d-b\overline{d})(c-c')-(b-d)(\overline{c}c'-c\overline{c'})}{(\overline{b}-\overline{d})(c-c')-(b-d)(\overline{c}-\overline{c'})} \\ &=\frac{(\frac{2x+b}{b}-b(2\overline{x}+\frac{1}{b})(c-\frac{b^2}{c})-(b-2x-b))(\frac{b^2}{c^2}-\frac{c^2}{b^2})}{(\frac{1}{b}-2\overline{x}-\frac{1}{b})(c-\frac{b^2}{c})-(b-2x-b)(\frac{1}{c}-\frac{c}{b^2}}\\ &=\frac{\frac{2xc}{b}-\frac{2c^2}{b}+\frac{2xc}{b}-\frac{2xb}{c}+\frac{2b^3}{c^2}-\frac{2xb}{c}+\frac{2xb^2}{c^2}-\frac{2xc^2}{b^2}}{-\frac{2c^2}{b^2}+\frac{2b^2}{c}} \end{align*}Now setting $p=p'$, a 22 term expansion yields the result.

Our aim is to show $\triangle ABC \sim \triangle DBQ$, as then there would exist a spiral similarity with center $B$ mapping $AC$ to $DQ$ and so by a well known lemma $AD \cap CQ$ would be on the circumcircle $\omega$ of $BQD$ (and also on the circumcircle of $ABC$). Since $\angle ACB = \angle DQB = 90^{\circ}$, we shall prove $\frac{BQ}{QD} = \frac{AC}{BC}$. Note that $\triangle PQD \sim \triangle PBQ$, so $\frac{PQ}{PB} = \frac{PD}{PQ} = \frac{DQ}{BQ}$, implying $\frac{PD}{PB} = \frac{BQ^2}{QD^2}$. Similarly from $\triangle ACH \sim \triangle CBH \sim \triangle ABC$ we obtain $\frac{AH}{BH} = \frac{AC^2}{BC^2}$. Hence we reduce to justifying $\frac{PD}{PB} = \frac{AH}{BH}$. But this follows by Menelaus' theorem on $\triangle ABD$ and the line $PXH$, where $X = AD \cap CH$ (note that $AX = XD$ by the problem conditions).

Let $AD$ intersect $(ABC)$ at $T$. Then, $$\angle DTB=\angle ATB=90,$$so $T$ lies on $(BDQ)$ as well. We wish to show that $CQ$ and $AD$ intersect at $T$. The key idea is that since $B$ lies on $(DQT)$ and $(ACT)$, this is equivalent to the fact that there is a spiral similarity at $B$ sending $QD$ to $CA$, or in other words, $$\triangle BQD\sim\triangle BCA\iff \angle QBD=\angle CBA.$$ Let $\angle QBD=\theta$ and let $F$ be the reflection of $A$ across $H$ which is also of course the foot from $D$ to $AB$. The idea is to focus on the tangency configuration to connect $\theta$ with the base triangle. We have $$\frac{PD}{PB}=(\frac{PD}{PQ})^2=\tan^2\theta$$but also $$\frac{PD}{PB}=\frac{HF}{HB}=\frac{HA}{HB}=\tan^2\angle CBA,$$so $\theta=\angle CBA$ as desired.

draw the other half of $\omega$, and let it intersect $AB$ at $R$ since $BD$ is a diameter of $\omega$, $DR$ is perpendicular to $AB$ and parallel to $CH$ let $CH$ intersect $AD$ at $S$, and since $AS=DS$ and $AHS$ is a right triangle, $AS=RS$ and $AH=HR$ we have that $PQ^2=PD*PB$ from PQ being a tangent, which leads to $PD/PB=PQ^2/PB^2=DQ^2/BQ^2$ but $PD/PB=HR/HB=AH/HB$ and $AH*HB=HC^2$ and $HC^2/HB^2=AH/HB$, so $HC/HB=DQ/BQ$, so $ACB$ and $DQB$ are similar the rest of this proof is trivial angle chasing

Denote E is midpoint of AD and $DF \perp BC$. We will prove that EC = EF. Denote K is the reflection of C across E $\Rightarrow$ AKDC is parallelogram so $KD \parallel AC$, and from $AC \perp BC$, $DF \perp BC$ we get that $AC \parallel DF$, so K, D, F lie on one line. From $\angle KFC = 90^{\circ}$ and CE = EK, it follows that EC = EF. Denote $(ABC) \cap (BD) = G$, $CQ \cap (BD) = Q$. From $\angle DFB = 90^{\circ}$, F lies on (BD), so $(BD) \cap BC = F$. We will prove that E, F, Q lie on one line. From $\angle DGB = 90^{\circ} = \angle ACB$ and $\angle ACB = \angle AGB$, we get that $\angle AGB = \angle DGB$. It follows that A, D, G lie on one line. Now $\angle BAC = 180 - \angle BGC = \angle BDQ$. It follows $\triangle ABC \sim \triangle DBQ$. From EC = EF and $\angle ACB = 90^{\circ}$, $\angle CFE = \angle ECF = \angle BAC = \angle BDQ = \angle BFQ$. We get that $\angle CFE = \angle BFQ$, so E, F, Q lie on one line. Now we use Pascal's theorem on QQGBDF and we get that the tangent through Q to (BD) passes through P. We are ready.

Redefine the problem in the following way: let $AD$ meet $\omega$ again at point $K,$ and let $CK$ meet $\omega$ again at point $Q.$ We wish to show that $PQ$ is tangent to $\omega.$ Note that $ACKB$ is clearly cyclic since $\measuredangle AKB = \measuredangle DKB = 90^\circ.$ Since $CQ$ and $AD$ intersect at point $K,$ there is a spiral similarity centered at $B$ mapping $DQ$ to $AC.$ Now, let the tangent at $C$ to $(ABC) = \Omega$ intersect $AB$ at point $Z,$ and let $X$ be the foot of the altitude from $D$ to $AB$ (which lies on $\omega$). It suffices to show that $P$ gets mapped to $Z$ by the spiral similarity, or equivalently, since $D$ gets sent to $A,$ $$\frac{BP}{DP} = \frac{BZ}{AZ}.$$However, since $\angle ACB = 90^\circ,$ we see that $(Z,H;A,B) = -1.$ Therefore, $$\frac{BZ}{AZ} = \frac{BH}{AH}.$$Since $CH$ bisects $AD,$ this implies that $AH = HX,$ so $$\frac{BZ}{AZ} = \frac{BH}{AH} = \frac{BH}{XH} = \frac{BP}{DP}$$since $DX \parallel PH$, and we conclude.

Tricky.... Let $X = AD \cap \omega$. Note from $\measuredangle AXB = 90 \Rightarrow X \in (ABC) $. If $Q' = CX \cap \omega$ then spiral similarity at $B$ sends $CQ' \rightarrow AD$. Which give us $\triangle BQ'D \sim \triangle BCA$. Therefore if $T \in BD$ such that $Q'T \perp BD$, then $HT \parallel AD$. Now $$(A,D;CH \cap AD , \infty{\overline{AD}})\stackrel{H}{=}(B,D;P,T)$$But as $\measuredangle DQ'B =90$ we have $PQ'$ tangent to $\omega$. Which give us $Q' = Q$.

[asy][asy] import geometry; size(8cm); point A = dir(180); point C = dir(110); point B = dir(0); triangle t = triangle(A,B,C); point H = foot(t.VC); point M = (-0.342, 0.1); // cos(110) was in radians for some reason point D = scale(2,A)*M; point P = intersectionpoint(line(B, D), line(C, H)); point F = intersectionpoint(parallel(H, line(A,D)), line(B,D)); point X = intersectionpoint(perpendicular(B, line(A,D)), line(A,D)); point Q = intersectionpoint(perpendicular(F, line(B,D)), line(C, X)); label("$A$", A, dir(180)); label("$B$", B, dir(0)); label("$C$", C, dir(110)); label("$H$", H, dir(225)); label("$M$", M, dir(120)); label("$D$", D, dir(225)); label("$P$", P, dir(180)); label("$F$", F, dir(270)); label("$X$", X, dir(0)); label("$Q$", Q, dir(90)); draw(A--B--C--cycle); draw(A--X); draw(C--H); draw(H--F); draw(B--P); draw(F--Q); draw(P--Q); draw(C--X, dashed+blue); draw(circle(B, D), red); draw(circle(A, B), red); [/asy][/asy] Throughout the proof I will refer to $\omega$ as if it is a circle (instead of a semicircle), but the point $Q$ will still be a unique point.We will prove the following: Claim: Let $AB$ be a diameter of a circle and $C$ a point on the circle. Let the foot from $C$ to $AB$ be $H$. Let $M$ be a point on $CH$ and take a point $N$ such that $\triangle BAM \stackrel{+}{\sim} \triangle BCN$. Let $D$ be the reflection of $A$ across $M$ and $Q$ the reflection of $C$ across $N$. Let $R$ be the intersection of $AM$ and $CN$. Let $BD$ intersect $CH$ at $P$. Prove that $PQ$ is tangent to $(BDQ)$. Proof. To see this implies the problem, note that $B$ would be the center of spiral similarity sending $AM$ to $CN$ and thus $AD$ to $CQ$. Let $R$ be the intersection of $AM$ and $CN$, so then this implies that $ACRB$ and $DQRB$ are cyclic and $\angle DQB = \angle ACB = 90$ implying that $(DQRB)$ is in fact $\omega$. Since $AD$ and $CQ$ intersect at $R$ which is on $(DQB)$ the result follows. $\square$ Let $E$ be the reflection of $A$ across $H$. Then $\angle DEB = 90$ so $E$ lies on $\omega$. Then note that \[ \angle PDQ = 180 - \angle BDQ = 180 - \angle BAC = \angle CEB \]Also, \[ DQ = AC \cdot \frac{BD}{AB} \implies \frac{DQ}{EB} = \frac{AC \cdot BD}{EB\cdot AB} \]But then, \[ \frac{AC\cdot BD}{EB\cdot AB} = \frac{AC\cdot PD}{HE\cdot AB} = \frac{AC \cdot PD}{AH \cdot AB} = \frac{PD}{AC} = \frac{PD}{CE} \]this readily implies $\triangle PDQ \sim \triangle CEB$, so then we find that $\angle PQD = \angle CBA = \angle QBD$ which implies the conclusion. $\blacksquare$

2015 ISL G3 Let $M$ be the midpoint of $AP$ and let $R$ be the intersection of the tangent at $C$ to $\omega$ and line $AB$. Firstly, we have that $(AD;M\infty )=-1$, we also know that $(RH;AB)=-1$, hence projecting through $P$ implies that $PT||AD$, by Thales Theorem we have $BD/BP=BA/BR$. So we easily get that $\triangle ABC \sim \triangle BDQ$ implying that $B$ is the Miquel point of $CQAD$ hence $CQ$ and $AD$ concur at $(ABC)$ and $(DBQ)$. hence we are done.

We claim that $\Delta ABC \stackrel{+}{\sim} \Delta DBQ$. Clearly $\angle ACB = 90^\circ = \angle DQB$. Let $A'$ be the reflection of $A$ across $H$. Clearly $CH // DA'$ since $CH$ bisects $AD$. Note that $\tan^2{\angle DBQ}=\left( \frac{\sin{\angle DBQ}}{\cos{\angle DBQ}} \right)^2=\left( \frac{\sin{\angle DQP}}{\sin{\angle QDP}} \right)^2=\left( \frac{PD}{PQ} \right)^2=\frac{PD}{PB}=\frac{A'H}{BH}=\frac{AH}{BH}=\left( \frac{CH}{BH} \right)^2 = \tan^2{\angle ABC}$, so $\angle ABC = \angle DBQ$ since both angles are acute. So $B$ is the centre of spiral similarity sending $AC$ to $DQ$. It follows that the intersection of $AD$ and $CQ$ lies on $(BDQ)$, so we are done. $\square$

Let $M$ be the midpoint of $AD$. Let $K$ be the second intersection point of $(ABC)$ and $\omega$. Let $N$ be the projection of $Q$ onto $BD$. Since $\angle BKD = \angle BKA$ we have $K \in AD$. Now let $CK \cap \omega \neq K = Q'$. It suffices to show $PQ'$ tangent to $\omega$ or that $Q'$ lies on the polar of $P$ wrt $\omega$. First, notice that $-1 = (A, D; M, \infty_{AD}) \overset{H} = (B, D; P; N' = H\infty_{AD} \cap BP)$. We can show that $N' = N$ which is sufficient as this implies that $QN$ is the polar of $P$. However it is clear that $B$ is the center of spiral similarity sending $\triangle BDQ \to BAC$. However $\triangle BNQ \sim \triangle BDQ \sim \triangle BAC$ so the spiral similiarity sending $\triangle BDQ \to \triangle BAC$ also sends $N$ to $H$. This implies $\triangle BHN \sim \triangle BAD \implies HN \parallel AD \implies N = N'$, done.

Let $E$ be the foot from $D$ to $\overline{HB}$ and let $F$ be the foot from $E$ onto $\overline{BC}$. Claim: We have $FHEB \sim QPDB$. (As a corollary of this, $E$, $F$ and $Q$ are collinear.)

to check that $HF$ is tangent to $(EFB)$. Indeed, due to cyclic quadrilateral $CHEF$, we have $\angle HFE = \angle HCE = \angle FBE$. Claim: We have $QCFB \sim DAEB$. Proof: Applying spiral similarity to our previous claim, we have $\triangle QFB \sim \triangle DEB$. Moreover, from $\triangle BAC \sim \triangle BEF$, we have $AE:EB = CF:FB$, so the claim follows. What we want to show follows from this similarity.

Let $X \in BD$ such that $HX \parallel AD$ Now, $(AD;M\infty) = -1 \implies (PX;BD) = -1$ $\implies QX \perp BD$ $HX \parallel AD \implies B$ is center of spiral similarity taking $\overline{DX}$ to $\overline{HA}$ As angles are preserved, $Q \longrightarrow C$ under the same transformation $\implies B$ is center of spiral sim taking $\overline{QD}$ to $\overline{CA} \implies AD \cap CQ \in \omega$

Used the 50\% hint on ARCH. Let $CQ$ and $AD$ meet at $R$, let $CA$ and $QD$ meet at $X$, and let $Z$ be the foot of the perpendicular from $Q$ to $DB$. Draw $(ABC)$. We will choose $Q$ to be the point so that the condition holds, and prove that $PQ$ is tangent to $\omega$. Note that $B$ is the Miquel point of $QCAD$. The idea is that $H$ on $ABC$ is the same position as $Z$ on $DQB$. Therefore, the spiral similarity sending $AC$ to $DQ$ should also send $H$ to $Z$. Then $AH$ is sent to $DZ$, so since they meet at $B$, this means $AD\parallel HZ$. Thus we have \[-1=(AD;M\infty)\stackrel{H}{=}(BD;PZ),\]so indeed we get the desired tangency. $\blacksquare$

[asy][asy] import graph; size(8cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -7.285601963776682, xmax = 6.892842353800001, ymin = -4.759777562678323, ymax = 8.25284545671791; /* image dimensions */ pen red = rgb(0.8,0,0); draw((-5,5)--(4,-1)--(-5,-1)--cycle, linewidth(0.7)); draw((-4.636311700472756,-1)--(-4.636311700472756,-0.6363117004727565)--(-5,-0.6363117004727565)--(-5,-1)--cycle, linewidth(0.7)); draw((-5,5)--(4,-1), linewidth(0.7) + red); draw((4,-1)--(-5,-1), linewidth(0.7)); draw((-5,-1)--(-5,5), linewidth(0.7)); draw((-4.03076923076923,0.4538461538461567)--(0.5384615384615381,-2.5), linewidth(0.7)); draw((-0.4615384615384608,-0.19230769230769074)--(4,-1), linewidth(0.7) + red); draw((-0.4615384615384608,-0.19230769230769074)--(0.5384615384615381,-2.5), linewidth(0.7)); draw((0.5384615384615381,-2.5)--(4,-1), linewidth(0.7)); draw(circle((1.76923076923077,-0.596153846153845), 2.267029571250247), linewidth(0.7)); draw((-0.4615384615384608,-0.19230769230769074)--(-0.46153846153846095,-1), linewidth(0.7)); draw((-0.46153846153846095,-1)--(0.5384615384615381,-2.5), linewidth(0.7)); draw(circle((-0.5,2), 5.408326913195984), linewidth(0.7)); draw((-4.03076923076923,0.4538461538461567)--(-0.4615384615384608,-0.19230769230769074), linewidth(0.7) + red); draw((-5,-1)--(1.871006874241814,-2.8608976951071563), linewidth(0.7)); draw((-5,5)--(-0.4615384615384608,-0.19230769230769074), linewidth(0.7) + red); draw((-0.4615384615384608,-0.19230769230769074)--(1.871006874241814,-2.8608976951071563), linewidth(0.7)); draw((-4.03076923076923,0.4538461538461567)--(-2.230769230769231,3.1538461538461537), linewidth(0.7) + blue); draw((-5,-1)--(-4.03076923076923,0.4538461538461567), linewidth(0.7)); /* dots and labels */ dot((-5,5),linewidth(3pt) + dotstyle); label("$A$", (-5.50283,4.926825870732575), NE * labelscalefactor); dot((4,-1),linewidth(3pt) + dotstyle); label("$B$", (4.184022278616064,-1.1937359447146654), NE * labelscalefactor); dot((-5,-1),linewidth(3pt) + dotstyle); label("$C$", (-5.70665,-1.3137469607038268), NE * labelscalefactor); dot((-2.230769230769231,3.1538461538461537),linewidth(3pt) + dotstyle); label("$H$", (-2.1422727070983205,3.315249370306691), NE * labelscalefactor); dot((-2.7307692307692304,2.4038461538461546),linewidth(3pt) + dotstyle); label("$D_{1}$", (-3.61,2.28658351897102), NE * labelscalefactor); dot((-0.4615384615384608,-0.19230769230769074),linewidth(3pt) + dotstyle); label("$D$", (-0.97301,-0.593), NE * labelscalefactor); dot((-4.03076923076923,0.4538461538461567),linewidth(3pt) + dotstyle); label("$P$", (-4.336759856614422,0.5721404334115693), NE * labelscalefactor); dot((0.5384615384615381,-2.5),linewidth(3pt) + dotstyle); label("$Q$", (0.2825336107293,-3.13461129711), NE * labelscalefactor); dot((1.871006874241814,-2.8608976951071563),linewidth(3pt) + dotstyle); label("$S$", (1.852379682255207,-3.2510676473860065), NE * labelscalefactor); dot((-0.46153846153846095,-1),linewidth(3pt) + dotstyle); label("$F$", (-0.8535738077951629,-1.5035822415016), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy] Let $AD\cap \omega=S$, $\omega\cap BC=F$ and $AD\cap CH=D_1$. We let $Q'=CS\cap \omega$. The problem is equivalent to proving $B$ is the center of spiral-similarity taking $DQ\to AC$ as $AD\cap CQ$ would lie on $\omega$. We apply Menelaus on $\triangle ADB$ to get $$\frac{AH}{HB}\cdot\frac{BP}{PD}\cdot\frac{DD_1}{D_1A}=1\implies \frac{AH}{HB}=\frac{PD}{BP}\iff \frac{AC}{CB}=\frac{DQ'}{Q'B}$$This gives $PQ'$ tangent to $\omega$ so $Q'\equiv Q$. Hence $\overline{C-Q-S}$ is collinear and we are done.

We can restate the question as follows: If $AD \cap (BD) = X$, and $CX \cap (BD) = Q'$, show $PQ'$ is tangent to $(BD)$. First notice that $\angle AKB = 90$, so $X \in (AB)$ too. Thus $B$ is the center of spiral similarity sending $AC \mapsto DQ'$. Let $E$ be the intersection of $BD$ with the line through $H$ parallel to $BC$, and $M$ be the midpoint of $AD$. Notice that $\angle DQ'B = 90$, and \[-1 = (AD:M\infty) \overset{H}{=} (BD,PE),\] so Apollonius gives us $\angle PQ'D = \angle DQ'E$. But our definition of $E$ also gives \[\frac{DE}{EB} = \frac{AH}{HB} \implies \angle DEQ' = \angle AHC = 90,\] so $\angle DQ'E = \angle DBQ'$. Thus $\angle PQ'D = \angle DBQ'$, so $PQ'$ is tangent to $(BD)$. $\blacksquare$