Consider $\Gamma$ the circumcircle of $\triangle ABC$, let $M\in \Gamma$ be the midpoint of the arc $BC$ containing $A$. Let $L=MA\cap CB$, $LO$ meets $\Gamma$ at $N,N'$ with $N$ in between $L,O$, and let $K'=MN\cap BC$ and $K''=MN'\cap BC$. Finally denote $\alpha=\angle B/2-\angle C/2$. 1- $\angle MLB=\angle AN'M=\alpha$, then $\angle MK'B=\angle MLB+\angle K'ML=\angle AN'M+\angle NN'M=\angle NNM$`, then $K'K''N'N$ is cyclic. 2- By power of point $K'K''MA$ is also cyclic with $K'K''$ as diameter and then is easy to check that $E$ is the midpoint of $K'K''$ and then $K'=K$. 3-$O$ is the orthocenter of $\triangle LME$ and then $LO\perp ME$, on the other hand $MODE$ is a parallelogram and then $DO$ perpendicular bisects $NN'$ and then $D$ is the circumcenter of $KK''N'N$. 4-With an easy angle work we get $\angle LAK=\angle N'K''K=\angle N'AM$ and $\angle KDN'=2\angle N'K''K $ wich implies $KDN'A$ is cyclic and then $N'=Q$. 5-The final step is just check that if $P'Q$ is tangent to $\Gamma$ with $P'\in BC$ then an easy angle chasing shows that $P'$ belongs to the circle through $K,D,Q,A$ and then $P=P'$ and we are done.

Has anyone bashed this successfully?

Similar to #1... Let $M$ be the midpoint of arc ${BAC}$. Let $K'$ be the reflection of $K$ over $E$ and denote $L=AM\cap BC$. Note that $OE\perp AM$, thus $OE$ is the perpendicular bisector of $AM \implies E$ is the circumcenter of $\odot(KAMK')$. Let $X=MK\cap \Gamma$ and $Y=MK'\cap \Gamma$. Clearly $XY$ passes through $O$, by radical axis we have that $XY$ also passes through $L$. Note that $O$ is the orthocenter of $\triangle{LME}$. thus $\angle{DOX}=90^{\circ}$. Because $MODE$ is a parallelogram we have that $DO=KE$ and $YO=DE$ implying that triangle $DOY$ and $KED$ are congruent, thus $DK=DY$. Denote $V=AK\cap \Gamma$. Now by Reim's on $\odot(AVYM)$ and $\odot(AKK'M)$ we have that $VY \mid\mid BC$, thus $AD$ bisects $\angle{KAY}$ and because $DK=DY$ by the incenter thing we have that $AKDY$ is cyclic, thus $Q=Y$. And because $VQ$ is parallel by Reim's theorem on $\odot(ADK)$ and $\Gamma$ we are done.

More general Problem: Let $ABC$ a triangle with circumcenter $O$, $D$ on the $A$-bisector and $E$ a point in the semi plane defined by $BC$ not containing $A$ such that $OE\parallel AD,DE\perp BC$.$K$ a point on $BC$ s.t. $KA=KE$; $(ADK)$ cuts $BC,(ABC)$ at $P(\ne K),Q(\ne A)$ .Prove that $PQ$ is tangent to $(ABC)$. Proof Let $G=(ABC)\cap AD$ ,since $OG\perp BC\implies OG \parallel DE $ thus $OGDE$ s parallelogram whence $ADEO$ is isoceles trapezoid hence if $F \in OE,I $ the circumcenter of $ADK$ then $O'F$ the bisector of $AD $ and of $OE$ besides $EO',EO$ are bisector of $KA,AQ$ whence $\angle KAD=\angle EO'F,\angle QAD=\angle OO'F$ but $AD$ is $A$-bisector then $\angle KAB=\angle CAQ$ more by easy angle chase we can prove the fact that any line cuting two secant circles determines two equal angles i.e $\angle KAB=\angle CQP$ therefore $\angle CAQ=\angle CQP$ which ends the proof . RH HAS

[asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(10cm); 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 = -12.1, xmax = 18.38, ymin = -11.19, ymax = 8.83; /* image dimensions */ pen wrwrwr = rgb(0.3803921568627451,0.3803921568627451,0.3803921568627451); draw((-5.76,4.23)--(-6.52,-2.23)--(1.84,-1.47)--cycle, linewidth(2)); draw(arc((-5.76,4.23),0.6,-36.86989764584403,-18.686422375158013)--(-5.76,4.23)--cycle, linewidth(2) + blue); draw(arc((-5.76,4.23),0.6,-114.89331207844296,-96.70983680775694)--(-5.76,4.23)--cycle, linewidth(2) + blue); /* draw figures */ draw((-5.76,4.23)--(-6.52,-2.23), linewidth(2)); draw((-6.52,-2.23)--(1.84,-1.47), linewidth(2)); draw((1.84,-1.47)--(-5.76,4.23), linewidth(2)); draw(circle((-2.5608108108108105,0.5789189189189193), 4.854400532388441), linewidth(2) + wrwrwr); draw((-2.5608108108108105,0.5789189189189193)--(-1.550060326128962,-1.7781873023753603), linewidth(2) + wrwrwr); draw((-2.5608108108108105,0.5789189189189193)--(-5.76,4.23), linewidth(2) + wrwrwr); draw((-2.5608108108108105,0.5789189189189193)--(-1.110563552339975,-6.612651814054214), linewidth(2) + wrwrwr); draw((-1.550060326128962,-1.7781873023753603)--(-1.110563552339975,-6.612651814054214), linewidth(2) + wrwrwr); draw(circle((-3.112762898121675,-1.0530266189587985), 5.9091652989393015), linewidth(2) + wrwrwr); draw((2.197796017683303,1.5385363326116739)--(2.7871319533288266,-1.383897095151925), linewidth(2) + wrwrwr); draw((-5.76,4.23)--(-1.550060326128962,-1.7781873023753603), linewidth(2) + wrwrwr); draw((-7.319417639304924,-0.3806984947738352)--(2.197796017683303,1.5385363326116739), linewidth(2) + red); draw((-5.76,4.23)--(-8.856272994005375,-2.442388454000489), linewidth(2) + wrwrwr); draw((2.197796017683303,1.5385363326116739)--(-5.76,4.23), linewidth(2) + wrwrwr); draw((-5.76,4.23)--(-1.110563552339975,-6.612651814054214), linewidth(2) + wrwrwr); draw((-8.856272994005375,-2.442388454000489)--(2.7871319533288266,-1.383897095151925), linewidth(2) + wrwrwr); draw((2.197796017683303,1.5385363326116739)--(-1.110563552339975,-6.612651814054214), linewidth(2) + red); draw((-8.856272994005375,-2.442388454000489)--(-1.110563552339975,-6.612651814054214), linewidth(2) + red); draw((-7.319417639304924,-0.3806984947738352)--(-1.110563552339975,-6.612651814054214), linewidth(2) + red); /* dots and labels */ dot((-5.76,4.23),dotstyle); label("$A$", (-6.02,4.49), NE * labelscalefactor); dot((-6.52,-2.23),dotstyle); label("$B$", (-6.34,-2.03), NE * labelscalefactor); dot((1.84,-1.47),dotstyle); label("$C$", (2.14,-1.31), NE * labelscalefactor); dot((-2.5608108108108105,0.5789189189189193),linewidth(4pt) + dotstyle); label("$O$", (-2.48,0.73), NE * labelscalefactor); dot((-1.550060326128962,-1.7781873023753603),linewidth(4pt) + dotstyle); label("$E$", (-1.48,-1.61), NE * labelscalefactor); dot((-1.110563552339975,-6.612651814054214),linewidth(4pt) + dotstyle); label("$D$", (-0.96,-7.09), NE * labelscalefactor); dot((2.197796017683303,1.5385363326116739),linewidth(4pt) + dotstyle); label("$Q' (Q)$", (2.28,1.69), NE * labelscalefactor); dot((-7.319417639304924,-0.3806984947738352),linewidth(4pt) + dotstyle); label("$Q''$", (-7.24,-0.23), NE * labelscalefactor); dot((-8.856272994005375,-2.442388454000489),linewidth(4pt) + dotstyle); label("$K$", (-9.36,-2.77), NE * labelscalefactor); dot((2.7871319533288266,-1.383897095151925),linewidth(4pt) + dotstyle); label("$P' (P)$", (3.04,-1.35), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy] Claim 1: $AOED$ is an isosceles trapezium. This is because $\angle EDA=\angle OAD=\frac{\angle B-\angle C}{2}$ and $OE//AD$. Construct the perpendicular at $O$ to $OD$, meeting the circumcircle of $\triangle ABC$ at $Q'$ and $Q''$. Let $Q'$ be the point on the same side as $C$ w.r.t. $OE$. Claim 2: $Q'=Q$. Firstly, $DQ'=DK$ as $DQ'^2=OQ'^2+OD^2=EA^2+OA^2=EK^2+ED^2=DK^2$. Furthermore, $2\angle Q'AC=\angle Q'OC=90^{\circ} -\angle COE-\angle EOD=90^{\circ}-(90^{\circ}-\angle B+\frac{\angle A}{2})-\angle OEA=\angle B-\frac{\angle A}{2}-\angle OEA$ $2\angle KAB=2\angle{EAK}-2\angle EAB=180^{\circ}-\angle AEB-\angle A-2\angle EAD=\angle B+\angle EAB-\angle A-2\angle EAD=\angle B-\frac{\angle A}{2}-\angle EAD=\angle B-\frac{\angle A}{2}-\angle OEA$ Thus $\angle KAB=\angle Q'AC\Rightarrow \angle KAD=\angle Q'AD$. This is only possible if $AQ'KD$ is concyclic. Thus $Q'=Q$. Let the tangent to circumcircle $\triangle ABC$ at $Q$ meet $BC$ at $P'$. Then $\angle QP'C=180^{\circ}-\angle QCP'-\angle CQP=180^{\circ}-\angle QAB-\angle QAC=180^{\circ}-\angle QAB-\angle KAB=180^{\circ}-\angle KAQ$, so $AKP'Q$ is concyclic, implying that $P'=P$.

Let $T$ be the intersection of $AB$ and the external angle bisector of $\angle BAC.$ Let $M$ be the intersection of $AD$ and $\odot(ABC).$ Let $N$ be the intersection of $\odot(ABC)$ and the external angle bisector of $\angle BAC.$ Notice that $OMDE$ is a parallelogram, it follows that $ED = OM = OA \implies AOED$ is an isosceles trapezoid. Moreover $T,A,O,E,D$ are concyclic with diameter $\overline{TD}.$ Now we consider $\sqrt{bc}$ inversion $\varphi$ with respect to $A.$ Clearly $\varphi(T) = N, \varphi(O) = O',$ where $O'$ is the reflection of $A$ with respect to line $BC.$ Then we deduce that $D': = \varphi(D) = NO' \cap AD$ and $\varphi(K) = KD' \cap \odot(ABC).$ Since $\varphi(Q)$ lies on $BC$ and $\varphi(\odot(AKD)) = \text{line } \varphi(K)D',$ so $\varphi(Q) = K.$ Now notice that $A,K,O', N$ are concyclic with center $E$(lies on $BC$), which implies that $\varphi(K) = Q, T, O$ are collinear, in other words $Q = TO \cap \odot(ABC).$ Therefore we have \begin{align*} PQ \perp OQ &\iff \angle TPQ + \angle PTQ = 90^\circ \iff 180^\circ - 2\angle KAD + \angle PTQ = 90^\circ\\ &\iff \angle TAD +\angle PTQ = 2\angle KAD \iff \angle TAK + \angle PTQ = \angle KAD\\ &\iff \angle TAK + \angle PTQ + \angle DAE = \angle KAE = \angle EKA\\ &\iff \angle PTQ + \angle DAE = \angle ATK \iff \overarc{$EO$} + \overarc{$DE$} = \overarc{$AE$}. \end{align*}

Attachments:

Disclaimer: Lengthy solution is coming! Let's apply an inversion centered at $A$ with radius $\sqrt{AB\cdot AC}$ followed by a reflection across the internal bisector of $\angle BAC$. The problem evolves to: Alternative problem. Let $ABC$ be a triangle with $AC>AB$. Let $O$ be the reflection of $A$ in $BC$. Let $E$ be a point on $(ABC)$ so that $(AEO)$ is tangent to the internal bisector of $\angle BAC$. This bisector meets $OE$ at $D$. Let $K$ be a point on $(ABC)$ such that $KA=KE$ and $K,\ C$ lie on the same side respect to $AE$. Line $KD$ intersects $(ABC)$ and $BC$ at $P$ and $Q$, respectively. Prove that $BC$ is tangent to $(APQ)$. The following known facts will help us to solve the problem. Claim 1. Let $ABCD$ be a cyclic quadrilateral with $P=\overline{AD}\cap \overline{BC}$, $Q$ its Miquel point, $Q'$ the reflection of $Q$ across $CD$ and $E=\overline{AC}\cap\overline{BD}$. Points $P,\ E$ and $Q'$ are collinear. Claim 2. Let $X,\ Y=\overline{CD}\cap (PAB)$. Then $QE$ is the internal bisector of $\angle XQY$. Let's turn to the solution of the alternative problem. It's easy to argue that $R=\overline{AM}\cap \overline{BC}$ (where $M$ is the midpoint of $\widehat{BAC}$) is the center of $(AOE)$ and with some angle-chasing we infer that $M,\ D,\ E$ and $O$ are collinear. Let $S$ be the antipode of $K$ on $(BAC)$ and redefine $Q$ to be the intersection point of $MS$ and $BC$. Let $T=\overline{MK}\cap \overline{BC}$. Since $ML\perp BC$ we get $$\angle KSM=\angle SML=\angle MTQ=\angle KTQ$$so $QSKT$ is cyclic. Since $(MSK)$ meets $MR$ again at $A$ then $A$ is the Miquel point of $QSKT$. By claims 1 and 2 $ME,\ KQ$ and $ST$ concur at $D$, so $KD$ passes through $Q$. Let $U=\overline{AL}\cap \overline{MQ}$. Note that $$\angle AUQ=90^\circ+\angle AMS=\angle AMK=\angle APQ$$, therefore $AUPQ$ is cyclic. Finally, observe that $$\angle QAU=90^\circ-\angle WAQ=90^\circ- \angle KTQ=90^\circ MSK=90^\circ-\angle QML=\angle MQC=\angle UQC$$so $BC$ is tangent to $(AUPQ)$, as required.

Let $Q'$ be the unique point such that $\triangle AKB \stackrel{+}{\sim} \triangle ACQ'.$ Clearly $Q' \in \odot(ABC).$ Furthermore, $PQ'$ is tangent to $\odot(ABC)$, because \[ \measuredangle PQ'A = \measuredangle PKA = \measuredangle BKA = \measuredangle Q'CA. \]It remains to prove that $Q' \equiv Q$, i.e. $A, K, D, Q'$ are concyclic. We shall in fact prove that $D$ is a midpoint of arc $\widehat{KQ'}$ on $\odot(AKQ').$ From the definition of $Q'$, we know that $AD$ bisects $\angle KAQ'.$ Thus, it remains to prove that $DK = DQ'.$ We prove the stronger result that $\triangle DKE \cong \triangle Q'DO.$ Let $M$ be the midpoint of arc $\widehat{BC}$ on $\odot(ABC).$ Clearly $OMDE$ is a parallelogram. In particular, $DE = OM = OQ'.$ Furthermore, we see that $AOED$ is an isosceles trapezoid. Hence, $EK = EA = OD.$ Finally, we calculate \[ \measuredangle DOQ = \measuredangle DOA + \measuredangle AOQ = \measuredangle DEA + 2\measuredangle ACQ = 90^{\circ} + \measuredangle KEA + 2\measuredangle AKE = 90^{\circ}. \]Thus $\triangle DKE \cong \triangle Q'DO$ by SAS congruence.

Let $W$ be the midpoint of the minor arc $BC$ and $J$ be the circumcenter of $(ADK)$. It’s sufficient to show that $\angle AQP= 180^{\circ}-1/2 \angle AOQ= \angle AOJ$. Note that $OEDW$ is a parallelogram, so $ED=OW=OA$, and since $OE \parallel AD$ we have that $AOED$ is an isosceles trapezoid. Since $J$ lies on the perpendicular bisector of $AD$, it lies on the axis of symmetry of $AOED$, so $\angle AOJ=\angle JED$. Finally, since $EJ$ is a perpendicular bisector of $AK$, $\angle JED=90^{\circ}+\angle JEK=180^\circ-\angle AKE=180^{\circ}-\angle AKP=\angle AQP$, as desired.

Only single point need to be constructed. As $OE\parallel AD$ and $\angle DAO = \angle ADE = \tfrac{\angle B-\angle C}{2}$, we get $AOED$ is isosceles trapezoid. Let $X$ be the center of $\odot(ADK)$ and let $\angle AEK = 2\theta$. As $AD, OE$ have the same perpendicular bisector, we get $XO=XE$. Moreover, $EX$ bisects $\angle AEK$. Thus we get \begin{align*} \angle XOE &= \angle XEO \\ &= \angle OED - 90^{\circ} - \theta \\ &= \angle AOE - 90^{\circ} - \theta \end{align*}Thus $\angle AOX = 90^{\circ} + \theta$. As $OX$ bisects $\angle AOQ$, we get $\angle AOQ = 2\theta$. Hence, we can conclude that $$\angle ABQ = \theta = \angle AKP = 180^{\circ}-\angle AQP$$so we are done.

Let $J$ be center of $(ADK)$ We have: $\angle{OAD} = \angle{CAD} - \angle{CAO} = \dfrac{\angle{BAC}}{2} + \angle{ACB} - 90^o = 90^o - \dfrac{\angle{BAC}}{2} - \angle{ABC}$ $= 90^o - \angle{(AD; BC)} = \angle{EDA}$ Combine with: $OE$ $\parallel$ $AD$, we have: $AOED$ is isosceles trapezoid Then: $\angle{JOA} = \angle{JED}$ But: $\angle{JOA} = 180^o - \angle{ABQ}$ and: $\angle{EJD} = 180^o - \angle{AKP}$ so: $\angle{ABQ} = \angle{AKP}$ or $PQ$ tangents $(O)$ at $Q$

Such an easy problem for China #3, but here is yet another solution. First, observe that $AEOD$ is an isosceles trapezoid. This follows from the fact that $O$ and $H$ are isogonal conjugates in $\triangle ABC$. Now we construct a point $Q'$ such that $OQ \perp OD$ and $OQ=OA$. The last condition implies $Q \in (ABC)$. We now show $A \in (Q'DK)$, which will yield $Q \equiv Q'$. For this let us denote the center of $(Q'DK)$ as $G$. From the construction of point $Q'$ we have $\triangle KED \cong \triangle DOQ \Rightarrow \triangle GEK \cong \triangle GOD \Rightarrow GE=GO \Rightarrow G$ lies on the perpendicular bisector of $OE$, and hence, of $AD$, therefore $GA=GD$, and $A$ does indeed lie on that circle, thus showing $Q \equiv Q'$. Now $\measuredangle AQP = \measuredangle AKP = \measuredangle GEK - 90^\circ=\measuredangle GED = \frac{1}{2}\widehat{AQ}$ in $(ABC)$, therefore showing $PQ$'s tangency to $(ABC)$. $\blacksquare$

Attachments:

From inscribed angle theorem, we can reduce the problem to proving $\angle{AKP}=\angle{ABQ}$. Now let $X$ be the center of $(AKD)$, and note $EX$ bisects $\angle{AEK}$ because $EX$ is the perpendicular bisector of $AK$; we desire to prove $\angle{ABQ}=90-\angle{XEK}$. Letting $M=AD\cap (ABC)$, we see $OMDE$ is a parallelogram, so $AO=OM=DE$, meaning $AOED$ is an isosceles trapezoid. Thus, the perpendicular bisector of $AD$ is the perpendicular bisector of $OE$, meaning $\triangle{XOE}$ is isosceles so $\angle{AOX}=\angle{XED}=90+\angle{XEK}$. Also, since $AQ$ is the radical axis of $(ABC), (ADK)$, we have $XO\perp AQ$, so since $AO=OQ$, $\angle{AOQ}=2(180-\angle{AOX})=2(90-\angle{XEK})$. Therefore, $\angle{ABQ}=90-\angle{XEK}$, as desired.

My solution although it might be similar to the above ones. We first prove the following $\underline{CLAIM}$ AOED is an isoceles trapezoid $\underline{Proof.}$Let $H$ be the projection of $A$ onto $BC$, since $AD$ bisects $\angle HAO$, $$\angle ADE=\angle HAD=\angle OAD$$Moreover, $OE\|AD$, this proves the claim. Now let $Q_1$ be the point on $(ABC)$ such that $OD\perp OQ_1$ and that $Q_1,C$ lies on the same side of $AD$. Let $Q_2$ be the antipode of $Q_1$. Let $AE$ meet $(ABC)$ again at $X$. Let $M$ be the arc-midpoint of arc $BC$ not containing $A$. From the claim, $ED=AO$ and $EK=EA=OD$, together with $\angle KED=\angle DOQ_1=90^{\circ}$, $$\triangle Q_1OD\cong\triangle DEK (1)$$ Meanwhile, $$\angle AOQ_2=\angle AOD-90^{\circ}=\angle AED-90^{\circ}=\angle OEK$$since $AO=OQ_2$, $AE=EK$, $$\triangle AOQ_2\sim\triangle AEK$$Hence, $$\angle ACQ_1=\angle AXQ_1=\angle AQ_2Q_1=\angle AQ_2O=\angle AKE=\angle AKB$$together with $\angle KBA=\angle AQ_1C$, $$\triangle AKB\sim\triangle ACQ_1$$Since $\angle BAD=\angle CAD$, we have $\angle KAD=\angle Q_1AD$ Moreover $DK=DQ_1$ from $(1)$, hence $D$ is the intersection of the angle bisector of $\angle KAQ_1$ and the perpendicular bisector of $KQ_1$, hence $A,K,D,Q_1$ are concyclic. This implies $Q_1=Q$ Now $$\angle AMQ=\angle ACQ_1=\angle AKB=\angle AKP=\angle ADP$$so $MQ_1\|DP$,by Reim's theorem, $PQ$ is tangent to $(ABC)$ since $A,D,M$ are collinear.

Diagram

$PQ$ is tangent to $(ABC)$ iff $\angle AQP = 180-\tfrac12 \angle AOQ=180-\angle ABQ$, which is equivalent to $\angle AKP=\angle ABQ$. But $\angle ABQ = \tfrac12 \angle AOQ$, so it suffices to show $\angle AOQ = 2\angle AKP$. The key is to involve $O'$, the circumcenter of $\triangle AKD$. Since $\angle ADE=\angle AMO=\angle OAM=\angle OAD$, we conclude $AOED$ is an isosceles trapezoid. And since $\triangle AKE$ is isosceles, and $O'$ is on the perp bisector of $AK$, we know $O'E$ bisects $\angle AEK$. Hence \[ \angle AOO' = \angle O'ED=90+\angle O'EK = 90+\tfrac12 \angle AEB. \]The angles around point $O$ sum to 360, so \[ \angle AOQ = 360-2\angle AOO' = 360-(180+\angle AEB) = 180-\angle AEB. \]So we want to show $2\angle AKP=180-\angle AEB$, and this follows since $\triangle AEK$ is isosceles.

Remarks

First China geo Hard geo problem nevertheless. $\textbf{Claim 01.}$ $AOED$ is an isosceles trapezoid. $\textit{Proof.}$ Notice that $\angle OED =\angle (AD,BC) + 90^{\circ} = 270 - \angle B - \frac{1}{2} \angle A$. But $\angle OAD = \frac{1}{2} \angle A + \angle B - 90^{\circ}$. Now, let $F$ be the circumcenter of $(AKD)$. Notice that $F$ lies on the perpendicular bisector of $AD$, so it lies on perpendicular bisector of $OE$ as well. Furthermore, $E$ and $F$ both lies on perpendicular bisector of $AK$. Now, \[ \angle FOE = \angle FEO = \angle OED - 90^{\circ} - \angle FEB = 90^{\circ} - \angle DAO - \angle FEB = \angle AOE - 90^{\circ} - \frac{1}{2} \angle AEB \]This gives us $\angle AOF = 90^{\circ} + \frac{1}{2} \angle AEB$. Notice that $FO \perp AQ$ since $AQ$ is the radical axis of both circles. Thus, \[ \angle ABQ = \frac{1}{2} \angle AOQ = 180^{\circ} - \angle AOF = 90^{\circ} - \frac{1}{2} \angle AEB = \angle AKP = 180^{\circ} - \angle AQP \]

Nice problem. Let $X$ be the circumcenter of pentagon $AKDPQ$. Note that $AOED$ is an isosceles trapezoid. We finish by angle chasing. Remark that $$\widehat{AQ} = \angle AOQ = 2 (180^\circ - \angle AOX) = 360^\circ - 2 \angle XED = 360^\circ - 2 (\angle XEK + \angle KED) = 360^\circ - 180^\circ - 2 \angle XEK = 180^\circ - \angle AEK = 2 \angle AKE = 2 \angle AKP = 2(180^\circ - \angle AQP),$$as desired.

mofumofu wrote: Let $O$ be the circumcenter of $\triangle ABC$($AB<AC$), and $D$ be a point on the internal angle bisector of $\angle BAC$. Point $E$ lies on $BC$, satisfying $OE\parallel AD$, $DE\perp BC$. Point $K$ lies on $EB$ extended such that $EK=EA$. The circumcircle of $\triangle ADK$ meets $BC$ at $P\neq K$, and meets the circumcircle of $\triangle ABC$ at $Q\neq A$. Prove that $PQ$ is tangent to the circumcircle of $\triangle ABC$. My Solution: Lemma: The angle between the altitude $AD$ and the circumdiameter $AL$ drawn from the vertex $A$ of $\angle ABC$ is equal to difference of angles at vertices $B$ and $C$ and is bisected by the angle-bisector $AX$ of $\angle A.$ Thus, the angle bisector of $\angle BAC$ is also an angle bisector of $\angle OAH.$ Proof. First, we note that $\angle AOC = 2B \implies \angle CAO = 90^{\circ}-\angle B = \angle DAB.$ Now, since $\angle XAB = \angle XAC,$ we get $\angle XAD = \angle XAO,$ i.e. $AX$ bisects $\angle DAL.$ At last, we note that \begin{align*} \angle DAL = 2(\angle XAB - \angle DAB) = 2\left(90^{\circ}-\left(\frac{\angle B+\angle C}{2}\right) - (90^{\circ} - \angle B) \right) = \angle B - \angle C \implies \angle DAX = \angle XAL = \frac{\angle B- \angle C}{2}.\quad\square \end{align*} [asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(7cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ real xmin = -4.201597952758092, xmax = 18.539052771683682, ymin = -6.245957701066118, ymax = 5.084331304245524; /* image dimensions */ pen evffff = rgb(0.8980392156862745,1.,1.); pen qqffff = rgb(0.,1.,1.); pen ffqqff = rgb(1.,0.,1.); pen xfqqff = rgb(0.4980392156862745,0.,1.); pen fsevff = rgb(0.9490196078431372,0.8980392156862745,1.); pair A = (3.477081478633382,3.5198622850388026), B = (3.,-1.), C = (10.,-1.), O = (6.5,0.9156768912914857), D = (7.3781928091451725,-4.989964655461024), K = (1.4581957838634185,-1.0059346992485405), P = (10.882865691493125,-1.), Q = (10.395608046851663,1.7782636388901194), F = (6.5,-3.074287764169539); filldraw(A--B--C--cycle, evffff, linewidth(0.8) + qqffff); filldraw(O--A--D--(7.378192809145173,-1.)--cycle, fsevff, linewidth(0.8) + xfqqff); /* draw figures */ draw(A--B, linewidth(0.8) + qqffff); draw(C--A, linewidth(0.8) + qqffff); draw(circle(O, 3.9899646554610246), linewidth(0.8) + green); draw(A--(7.378192809145173,-1.), linewidth(0.8) + ffqqff); draw(circle((6.170147691512336,-0.39466631832606186), 4.7514355529157575), linewidth(0.8) + qqffff); draw(O--Q, linewidth(0.8) + blue); draw((7.378192809145173,-1.)--K, linewidth(0.8) + ffqqff); draw(O--F, linewidth(0.8) + xfqqff); draw(B--Q, linewidth(0.8) + blue); draw(O--A, linewidth(0.8) + xfqqff); draw(A--D, linewidth(0.8) + xfqqff); draw(D--(7.378192809145173,-1.), linewidth(0.8) + xfqqff); draw((7.378192809145173,-1.)--O, linewidth(0.8) + xfqqff); draw((6.170147691512336,-0.39466631832606186)--(7.378192809145173,-1.), linewidth(0.8) + green); draw((5.427637143889277,-0.7350511852111106)--(6.939096404572587,-0.04216155435425717), linewidth(0.8) + green); draw(A--Q, linewidth(0.8) + yellow); draw(Q--P, linewidth(0.8) + blue); draw(A--K, linewidth(0.8) + yellow); draw(K--D, linewidth(0.8) + yellow); draw((6.9363447627425225,2.6490629619644612)--(6.170147691512336,-0.39466631832606186), linewidth(0.8) + green); draw((7.378192809145173,-1.)--P, linewidth(0.8) + qqffff); /* dots and labels */ dot(A,linewidth(4.pt)); label("$A$", (3.565455287632233,3.683058812422529), NE * labelscalefactor); dot(B,linewidth(4.pt)); label("$B$", (3.0850190047214916,-0.8410495183202817), NE * labelscalefactor); dot(C,linewidth(4.pt)); label("$C$", (10.07136328538186,-0.8410495183202817), NE * labelscalefactor); dot(O,linewidth(4.pt)); label("$O$", (6.5882002342789825,1.0806956133226824), NE * labelscalefactor); dot((7.378192809145173,-1.),linewidth(4.pt)); label("$E$", (7.448981907827395,-0.8410495183202817), NE * labelscalefactor); dot(D,linewidth(4.pt)); label("$D$", (7.448981907827395,-4.824667030788509), NE * labelscalefactor); dot(K,linewidth(4.pt)); label("$K$", (1.543619263716195,-0.8410495183202817), NE * labelscalefactor); dot(P,linewidth(4.pt)); label("$P$", (10.972181315839501,-0.8410495183202817), NE * labelscalefactor); dot(Q,linewidth(4.pt)); label("$Q$", (10.471726854474145,1.9414772868710932), NE * labelscalefactor); dot(F,linewidth(4.pt)); label("$F$", (6.5882002342789825,-2.9229400776001593), NE * labelscalefactor); dot((6.170147691512336,-0.39466631832606186),linewidth(4.pt)); label("$O'$", (6.247891200550541,-0.24050416468185545), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */[/asy][/asy] So, we have $\angle OAD = \tfrac{\angle B- \angle C}{2}.$ Claim: $AOED$ is an isosceles trapezium. Proof. Let the $\angle A$-bisector intersect minor $\widehat{BC}$ at $F$, we know that $OA=OF \implies \angle OFA = \tfrac{\angle B- \angle C}{2}.$ Since $DE\perp BC \implies DE \parallel OF$, and as $\{D\} \in \angle A$-bisector, we get $\angle ADE = \tfrac{\angle B- \angle C}{2}.$ And as $AD \parallel OE$, the conclusion follows.$\qquad\square$ Let $O'$ denote the center of $(ADK)$ and $\angle AEK = 2\alpha.$ As $AD, OE$ have the same perpendicular bisector, so we get $OO'=OE.$ Also, $O'E$ bisects $\angle AEK$, so, $$\angle AOO' =\angle O'ED = 90^{\circ} + \alpha.$$Since, $OO'$ bisects $\angle AOQ$, $$\implies \angle AOQ = 180^{\circ}-2\alpha \implies \angle ABQ = 90^{\circ}-\alpha = \angle AKP = 180^{\circ} - \angle AQP.\qquad\blacksquare$$

Let $F$ be the foot of the altitude from $A$ to $BC$. Then $\measuredangle DAO=\measuredangle FAD=\measuredangle EDA$. Thus $AOED$ is an isosceles trapezoid. Then let $G$ be the center of $(AQPDK)$; note that $F$ is on the perpendicular bisector of $AD$ and hence also on the perpendicular bisector of $OE$. Also note that because $\triangle AEK$ is isosceles $EG$ bisects $\angle AEK$. Also it is evident that $OG$ bisects $\angle AOQ$. Then \[\angle PBA = \frac12 \angle QOA=\frac12(360^\circ - 2\angle AOG)=180^\circ - \angle GED = 90^\circ - \frac12 \angle AEK=\angle AKP=180^\circ - \angle AQP,\]and we're done.

Let $R$ be the circumcenter of $\triangle AKD$, because $EK=EA$ so we have $RE\perp AK$, $RO\perp AQ$ notice that $\angle AKP=90^{\circ}-\angle REK=180^{\circ}-\angle RED=180V-\angle ROA=\angle ABQ$ Therefore, we have $\angle ABQ=\angle AKP=180^{\circ}-\angle AQP$ Which means $PQ$ is tangent to the circumcircle of $\triangle ABC$. IS THIS ANSWER CORRECT? SEEMS TOO SIMPLE AS A P3!

Lemma: Consider a cyclic, convex quadrilateral $XYZW$ with circumcenter $O$ and Miquel Point $M$. Then, $MXOZ,MYOW$ are cyclic. Proof: Let $\omega= (XYZW), T= XZ \cap YW, U= XY \cap ZW, V=XW \cap YZ$. It's well known that $UV= \Pi_{\omega}(T)$ and $M \in UV$. Furthermore, $O,T,M$ are collinear. Now, invert about $\omega$. $\square$ Let $R= AK \cap \Gamma, R \neq A$. Redefine $Q$ to be the point on $\Gamma$ such that $QR \parallel BC$. We will prove that $Q \in (AKD)$, which solves the problem. Let $S= OQ \cap \Gamma, S \neq Q$ and let $K'$ be the reflection of $K$ WRT $E$. Clearly, $EK=EA=EA' \implies \angle KAK'= 90º=\angle SAQ (*)$, since $SQ$ is a diameter of $\Gamma$. Now, $\angle ASQ= \angle ARQ= \angle AKK' \implies $ from $(*)$, $\Delta ASQ \sim \Delta AKK' \implies A$ is the Miquel point of $SKK'Q (\spadesuit)$, i.e., the center of the spiral similarity mapping $SQ \mapsto KK'$. Let $L$ be the midpoint of $\widehat{BAC}$ and let $M$ be the midpoint of $\widehat{BC}$. Since $OE \parallel MD; ED,OM \perp BC \implies EDMO$ is a parallelogram $\implies ED=OM=OL \implies EDOL$ is a parallelogram $\implies EL=OD (\heartsuit)$. Moreover, since $ED=OM=OA \implies DEOA$ is an isosceles trapezoid $\implies OD=EA$ from $(\heartsuit)$, $EA=EL \implies L$ lies on $(KAK')$. Therefore, since $L \in \Gamma=(SAQ), (KAK') \implies$ from $(\spadesuit)$, $L=SK \cap QK' \quad (\star)$. Let $\Omega= (SKQ) \implies$ by Reim's Theorem on $\Gamma, \Omega$ with the lines $SL,QL$, the lines $LL, K\{LQ \cap \Omega\}$ are parallel. However, $LL \parallel KK' \implies K'= LQ \cap \Omega \implies K' \in \Omega \implies KSQK' $ is cyclic. Now, observe that the center of $\Omega$ must lie on the perpendicular bisector of $KK'$, which is $ED$, and since $L= KS \cap K'Q$ and $(\spadesuit)$, the center of $\Omega$ must lie on the line through $A$ perpendicular to $AL$, which is $AD \implies D$ is the center of $\Omega$. Therefore, from the lemma, $Q \in (AKD)$. To finish, apply Reim's Theorem on $\Gamma, (AKD)$, with the lines $QQ, ARK \implies PQ$ is tangent to $\Gamma$, as desired. $\blacksquare$

We pick the circumcenter $O'$ of $\triangle ADK$. It is easy enough to prove the following conclusions in order: 1. AOED is an isosceles trapezoid, so we pick its circumcentre $O_1$ 2. $O_1O'\perp AD$ thus $\perp OE$ so $O'O=O'E$. 3. $O'A=O'K$ and $EA=EK$ implies $O'E\perp AK$ while naturally $O'O\perp AQ$, so easy angle chase gives $AD$ bisects $\angle KAQ$, which means $\angle KAB=\angle CAQ$. 4. With $\angle ABK=\angle AQC$ we have $\angle PQA=180^{\circ}-\angle PKA=180^{\circ}-\angle ACQ$ thus done.

Let $M$ be the midpoint of arc $BC$. Then $OA = OM$, and $MOED$ is a parallelogram, so $AOED$ is an isosceles trapezoid, and so $KE = AE = DO$. Now let $Q'$ be the point on $(AKD)$ such that $DK = DQ'$. Then \[ \angle ODQ' = \angle ADQ' - \angle ODA = \angle AKD - \angle KAD - \angle DAE = \angle AKD - \angle KAE = \angle AKD - \angle AKE = \angle EKD, \]so $\triangle KED \cong \triangle DOQ'$, and hence $OQ' = ED = OM$. Therefore, $Q'$ lies on $(ABC)$ and $Q' = Q$. Now note that \[ \angle CQP = \angle KAB = \angle KAD - \angle BAD = \angle DAQ - \angle DAC = \angle CAQ, \]so $\overline{PQ}$ is tangent to $(ABC)$ as desired.

wow this p3 i solved on my own Let M be the midpoint of arc BC, which lies on AD, meaning DEOM is a parallelogram, AOM is a isosceles triangle, AOED is an islscelestrapezoid, and let O' be the circumcenter of (AKDPQ); since O',E both lie on perp. bisector of AK implies $$90+90-AKP=90+O'EK=O'ED=O'OA\stackrel{radax}{=}90+OAQ=180-ABQ\implies180-AQP=AKP=ABQ,$$as desired. $\blacksquare$

Let $H$ be the circumcenter of $(AKD)$. Let $AD$ meet $BC$ at $F$ and meet $\Gamma$ at $G$. Let $J$ be the midpoint of $AK$ and let $L$ be the midpoint of $AQ$. Claim: $AOED$ is an isosceles trapezoid. Proof: Note that $\angle FAO = \angle OGA = \angle EDG$, as $OGDE$ is a parallelogram. Now clearly we have $\angle HOE = \angle HEO$ as $H$ lies on the perpendicular bisector of $AD$. Clearly $E$, $H$, $J$ are collinear from $AE = EK$ and so are $H$, $O$, $L$ from radax. Claim: $\triangle ACQ \sim \triangle AKB$ Proof: First note that clearly $\angle AQB = 180 - \angle ABC = \angle ABK$. Now we will show that $\angle ACQ = \angle AKB$. To see this note that we have, \begin{align*} \angle ACB &= \angle AOL\\ &= 180 - \angle AOH\\ &= 180 -\angle HED\\ &= 90 - \angle JEK\\ &= \angle JKE, \end{align*}hence proved. Now note that the result follows. Now we finish. Clearly it suffices to show that $\angle PQC = \angle QAC$. To see this note that we have, \begin{align*} \angle CQP &= 180 - \angle QCP - \angle QPC\\ &= 180 - (180 - \angle QCK) - (180 - \angle KAQ)\\ &= (360 - \angle AKC - \angle AQC) - 180\\ &= 180 - \angle ACQ - \angle AQC\\ &= \angle CAQ \end{align*}as desired. I guess the second claim is a bit uneccesary

Define $M$ as the midpoint of arc $BC$, and $X$ as the center of $(ADK)$. Then $ADEO$ is an isosceles trapezoid, with $AO = OM = ED$. $XO$ forms the perpendicular bisector of $QA$. We notice that $\triangle ABK \sim \triangle AQC$, as $\measuredangle KBA = \measuredangle CQA$ and \[\measuredangle ACQ = \measuredangle APX = \measuredangle XED = \measuredangle AKB.\] This essentially finishes, as we deduce \[\measuredangle PQC = \measuredangle PQA + \measuredangle AQC = \measuredangle PKA + \measuredangle AQC = \measuredangle QCA + \measuredangle AQC = \measuredangle QAC. \quad \blacksquare\] Remark: Hilariously, there is also an inversive finish. Force overlaying swaps $B$, $C$ as well as $Q$, $K$, and sends $P \mapsto QK \cap (ABC)$. Thus we need to show $(AKP^*)$ is tangent to $BC$, which is fairly simple.

Let $M,N$ be the midpoints of arcs $BC,BAC$ respectively. $AN\cap BC=T,NK\cap (ABC)=S$. Note that $DE=OM=AO$ so $A,O,D,E$ are concyclic. Also $T$ lies on this circle. Claim: $T,S,O$ are collinear. Proof: Note that $OE\parallel AD\perp AN$ hence $E$ lies on the perpendicular bisector $AN$ or $EK=EA=EN$. \[\measuredangle SOA=2 \measuredangle SNA=2 \measuredangle KNA=\measuredangle KEA\]And combining this with $EA=EK$ and $OS=OA$ implies that $A$ is the center of spiral homothety sending $SO$ to $KE$. Since $A$ is the miquel point of $SOEK$, $(AOE)$ passes through $BC\cap SO$, observe that this point must be $T$ hence $T,S,O$ are collinear.$\square$ Claim: $T,O,Q$ are collinear. Proof: Let $TO\cap (ABC)=S,Q'$. We want to show that $Q',A,D,K$ are concyclic. Let $DO\cap NS=L$. \[\measuredangle AKL=\measuredangle ATO=\measuredangle ADO=\measuredangle ADL\]Also \[\measuredangle Q'LD=\measuredangle QLO=\measuredangle Q'NO=\measuredangle Q'NM=\measuredangle Q'AD\]Thus, $A,K,D,L,Q$ are concyclic.$\square$ Claim: $PQ$ is tangent to $(ABC)$. Proof: Let $NQ\cap BC=R$. Under the inversion centered at $N$ with radius $NB$, $TSQ$ swaps with $(AKNR)$. \[\measuredangle PQA=180-\measuredangle AKQ=180-\measuredangle AKR=\measuredangle RNA=\measuredangle QNA\]Hence $PQ$ is tangent to $(ABC)$ as desired.$\blacksquare$