Let $ BO_B\cap CO_C = Y$. We have $ \angle CBO_B = 90^{\circ} - \angle DAB$ and $ BCO_C = 90^{\circ} - \angle DAC$ hence $ \angle BYC = 180^{\circ} - (90^{\circ} - \angle DAB) - (BCO_C = 90^{\circ} - \angle DAC = \angle BAC$ which means that point $ Y$ lies on the circumcircle of triangle $ ABC$ (1). A simple angle chase shows that $ \angle BAO_B = \angle CAO_C$ which implies $ \angle O_BAO_C = \angle BAC$ and point $ Y$ lies on the circumcircle of triangle $ AO_BO_C$ (2). From (1) and (2) we have $ AY$ - radical axis of circles $ (ABC)$ and $ (AO_BO_C)$, $ O_BO_C$ - radical axis of circles $ (BCO_CO_B)$ and $ (AO_BO_C)$ and $ BC$ - radical axis of $ (ABC)$ and $ (BCO_CO_B)$ and these three lines are concurrent in point $ R$. Let $ AH\cap BC = T$ and $ O_BO_C\cap AD = Q$. We know that $ AD$ is radical axis of circles $ (ABD)$ and $ (ACD)$ hence $ O_BO_C\perp QR$ and $ AT\perp RT$ which implies that $ RTQA$ is cyclic quadrilateral and $ \angle DAH = \angle RAT = \angle DRQ$ (3). We know that $ AX\perp AY$ (4) (it was posted on the forum, I'll write proof later). A simple angle chase shows that $ RBO_BA$ is cyclic quadrilateral and $ BO_B = AO_B$ hence $ QR$ is angle bisector of $ \angle DRA$ which combined with $ AD\perp QR$ implies $ AR = DR$ and $ \angle RDA = \angle DAR$ (5). Let $ AX\cap QR = S$. Using (4) and (5) we get $ \angle RDA = \angle DAR = \angle RSA$ (because $ \triangle RQA$ and $ \triangle RAS$ are similar) hence $ RDSA$ is cyclic. Using this and (3) we get $ \angle DAX = \angle DAS = \angle DRS = \angle TRQ = \angle TAQ = \angle HAD$ qed. We can also prove that $ BO_C$, $ CO_B$ and $ AX$ are concurrent.

Let $ F \in BC$ be foot of $ AH.$ Let $ A', B', C', D', F'$ be midpoints of $ BC, CA, AB, AD, AF.$ $ O_BO_C$ is perpendicular bisector of $ AD$ $ \Longrightarrow$ $ \angle AO_BO_C = \angle ABD = \angle ABC$ and $ \angle AO_CO_B = \angle ACD = \angle ACB$ $ \Longrightarrow$ $ \triangle AO_BO_C \sim \triangle ABC \sim \triangle AC'B'$ for any $ D \in BC.$ Let $ K, M$ be midpoints of $ O_BO_C, C'B'$ and $ AD'KL, AF'MN$ rectangles; $ KL, MN$ are perpendicular bisectors of $ O_BO_C, C'B',$ respectively. Since $ D', F'$ are A-altitude feet of $ \triangle AO_BO_C \sim \triangle AC'B',$ these 2 rectangles are similar $ \Longrightarrow$ $ \triangle ALN \sim \triangle AD'F'$ are similar by SAS and $ LN \perp (AN \parallel B'C').$ Consequently, foot $ L$ of perpendicular $ AL$ to $ KL$ from $ A$ is on midparallel $ MN$ of $ AH \parallel OA'.$ This means that the perpendicular bisector $ KL$ of $ O_BO_C$ is tangent to a parabola with focus $ A,$ vertex tangent $ MN$ and directrix $ OA'.$ If $ KL$ cuts the directrix $ OA'$ at $ X,$ the parabola tangent $ KL$ bisects the angle $ \angle OXA.$ Since $ AD \parallel KL$ $ \Longrightarrow$ $ AD$ bisects the angle $ \angle XAH.$ This is true regardless of $ BCO_CO_B$ being cyclic or not. But if it happens to be cyclic, the intersection $ X$ of the perpendicular bisectors $ OA', KL$ of $ BC, O_BO_C$ is its circumcenter.

I will post the solution later in mind

Denote X’ is the intersection of B’C’ and ObOc ⊿ABOb∽⊿ACOc => ∠ObBC’=∠OcCB’ => C’OcB’Ob is an isosceles trapezoid => ObOc=B’C’ ⊿AC’B’∽⊿ACB ⊿AOcOb∽⊿ACB => B’C’/BC=AB’/AC= ObOc/BC=AOc/AC => AB’=AOc Analogously, we have AOb=AC’ AX’ is the bisector of ∠C’AOc On the other hand, we have AX is the bisector of ∠C’AOc Thus A,X,X’ are collinear By angle chasing ∠A’AB+∠C’AC=90 which indicates A’,O,X’’are collinear Therefore ∠OX’’A=∠OAX=∠ABA’=∠DAH (1) Notice ∠OAX=∠X’AC’-∠OAB=∠X’Oc-∠OcAD=∠DAX (2) Form (1)(2) we get the result

plane geometry wrote: Denote $ X'$ is the intersection of $ B'C'$ and $ O_bO_c$ $ \triangle ABO_b\sim\triangle ACOc \Rightarrow \angle O_bBC' = \angle O_cCB' \Rightarrow C'O_cB'O_b$is an isosceles trapezoid $ \Rightarrow O_bO_c = B'C'$ ${ \triangle AC'B'\sim \triangle ACB \triangle AO_cO_b\sim \triangle ACB \Rightarrow \frac {B'C'}{BC} = \frac {AB}{AC} = \frac {O_bO_c}{BC} = \frac {AO_c}{AC } } \Rightarrow AB = AO_c$ Analogously, we have $ AO_b = AC'$ $ AX'$ is the bisector of $ \angle C'AO_c$ On the other hand, we have $ AX$ is the bisector of $ \angle C'AO_c$ Thus $ A,X,X'$ are collinear By angle chasing $ \angle A'AB + \angle C'AC = 90^0$ which indicates $ A',O,X''$ are collinear Therefore$ \angle OX''A + \angle OAX + \angle ABA' + \angle DAH (1)$ Notice $ \angle OAX = \angle X'AC' - \angle OAB = \angle X'AO_c - \angle O_cAD = \angle DAX (2)$ From$ (1)(2)$ we get the result Very nice solution! Dear plane geometry

MellowMelon wrote: Let $ ABC$ be an acute triangle. Point $ D$ lies on side $ BC$. Let $ O_B, O_C$ be the circumcenters of triangles $ ABD$ and $ ACD$, respectively. Suppose that the points $ B, C, O_B, O_C$ lies on a circle centered at $ X$. Let $ H$ be the orthocenter of triangle $ ABC$. Prove that $ \angle{DAX} = \angle{DAH}$. Zuming Feng. Dear mathlinkers! How can we construct $ D$ on $ BC$ such that $ O_B,O_C,B,C$ are on a circle where $ O_B, O_C$ is the circumcenters of triangles $ ABD$ and $ ACD$, respectively?

mathVNpro wrote: Dear mathlinkers! How can we construct $ D$ on $ BC$ such that $ O_B,O_C,B,C$ are on a circle where $ O_B, O_C$ is the circumcenters of triangles $ ABD$ and $ ACD,$ respectively? The point $ D$ is not constructible (in general). Calculate the angles $ \angle CAD$ and $ \angle BAD$ in terms of the angles of $ \triangle ABC,$ then realize that the construction is equivalent to the angle trisection.

Luis González wrote: ...the construction is equivalent to the angle trisection. Calulate angle $ \theta = \angle DAH$: $ \angle O_BBC = \theta + \angle ABC$ and $ \angle BCO_C = \angle BCA - \theta ,$ $ \angle O_CO_BB = \pi - (\angle ABC + 2 \theta )$ and $ \angle CO_CO_B = \pi - (\angle BCA - 2 \theta).$ $ BCO_CO_B$ is cyclic $ \Longleftrightarrow$ $ \angle O_BBC + \angle CO_CO_B = \pi$ $ \Longleftrightarrow$ $ \theta = \frac {_1}{^3}(\angle BCA - \angle ABC).$ Even though $ D$ is not constructible for a given $ \triangle ABC,$ you can still construct some $ \triangle ABC$ with $ BCO_CO_B$ cyclic, given $ \triangle ABD.$ (If $ F \in BD$ is A-altitude foot, pick $ D,$ so that $ \angle ABD + 3 \angle DAF < \frac{_{\pi}}{^2}.$)

We know that $ AX$ is the bisector of $ \angle BAO_c$ and $ \Delta AO_bO_c\sim \Delta ABC$ On the other side, denote $ T$ the intersection of $ BO_c$ and $ CO_b$. We have $ \frac {O_bT}{O_cT} = \frac {BO_b}{CO_c}$ $ \Rightarrow \frac {O_bT}{TC} = \frac {R_b}{R_c}.\frac {O_cT}{TC} = \frac {R_b}{R_c}.\frac {O_bO_c}{BC}= \frac {R_b}{AC} = \frac {AO_b}{AC}$ So $ AT$ is the bisector of $ \angle O_bAC$ or $ \angle BAO_c$ We obtain $ AX, BO_b, CO_c$ are concurrent. Applying Tri-Ceva theorem for triangle $ O_bO_cX$ we have: $ \frac {sin \angle O_cXT}{sin \angle O_bXT}.\frac {sin \angle XO_bT}{sin \angle O_cO_bT}.\frac {sin \angle O_bO_cT}{sin \angle XO_cT} = 1$ Note that $ XO_b = XO_c$ so $ \frac {sin \angle O_cXT}{sin \angle O_bXT} = \frac {PO_c}{PO_b}$ And $ \frac {sin \angle XO_bT}{sin \angle O_cO_bT}.\frac {sin \angle O_bO_cT}{sin \angle XO_cT} = \frac {sin \angle TCB}{sin \angle TBC}.\frac {cos \angle O_bBD}{cos \angle O_cCD} = \frac {TB}{TC}.\frac {cos \angle O_bBD}{cos \angle O_cCD}$ $ = \frac {R_b}{R_c}.\frac {cos \angle O_bBD}{cos \angle O_cCD} = \frac {BD}{CD}$ Therefore $ \frac {PO_c}{PO_b} = \frac {CD}{BD}$ We get $ \Delta ABD\sim \Delta AO_bP$ then $ \angle BAD = \angle O_bAP \Rightarrow \angle DAX = \angle BAO_b = \angle DAH$

Quote: Let $ ABC$ be an acute triangle. Point $ D$ lies on side $ BC$. Let $ O_B, O_C$ be the circumcenters of triangles $ ABD$ and $ ACD$, respectively. Suppose that the points $ B, C, O_B, O_C$ lies on a circle centered at $ X$. Let $ H$ be the orthocenter of triangle $ ABC$. Prove that $ \angle{DAX} = \angle{DAH}$. Zuming Feng. We have $ \angle AO_BB = 2(180^{\circ} - \angle ADB) = 2\angle ADC = \angle AO_CC$. In the other hand, since $ \triangle AO_BB$ and $ \triangle AO_CC$ are all issoceles triangles respectively with respect to vertexes $ O_B,O_C$. Hence it exists a spiral similarity through center $ A$, let denote this transformation by $ f_A$ such that $ f_A:$ $ \triangle AO_BB\mapsto \triangle AO_CC$ $ (*)$. Thus $ f_A:$ $ O_B\mapsto O_C$ and $ B\mapsto C$. Therefore, it also exists a spiral similarity through $ A$, let denote this by $ g_a$ such that $ g_A:$ $ O_b\mapsto B$, $ O_c\mapsto C$ $ \Longrightarrow$ $ O_BO_C\mapsto BC$, which implies that $ \triangle AO_BO_C\mapsto \triangle ABC$ $ (**)$. From $ (*)$ and $ (**)$, we have $ (O_BO_C,BC) = (AO_c,AC)$ and $ (O_BB,O_CC) = (AB,AC)$. Thus, if $ P\equiv BO_B\cap CO_C$ and $ Q\equiv O_BO_C\cap BC$ then we have $ \{A,O_C,C,Q\}$ and $ \{P,B,C,A\}$ are two sets of concyclic points, which leads to the fact that $ A$ is the Miquel point with respect to the complete quadrilateral $ (PB,PC,QO_B,QB)$. Note that $ OO_BO_CC$ is concyclic $ \Longrightarrow$ $ A,P,Q$ are collinear. Further, if $ G\equiv CO_B\cap BO_C$ then $ PQ$ is the polar of $ G$ with respect to $ (O_A)\equiv (BO_BO_CC)$, $ O,A,G$ are collinear and $ \overline {OGA}\perp PQ$ at $ A$ $ \Longrightarrow$ $ \angle PAO_A$ $ = \angle O_AAQ$ $ = 90^{\circ}$. In the other hand, we have $ \angle DAH_A = \angle DAC - \angle H_AAC = \angle DAC - (90^{\circ} - \angle ACD)$. Also, $ \angle DAO_A = \angle PAD - \angle PAO_A$ $ = \angle PAD - 90^{\circ}$. Therefore, in order to prove that $ \angle DAH_A = \angle DAO_A$, we need to prove that $ \angle PAD = \angle DAC + \angle ACD$ $ = 180^{\circ} - \angle ADQ$ $ = 180^{\circ} - \angle QAD$, which equivalent to $ \angle ADQ = \angle QAD$, which is obviously true since $ \overline {O_BO_CQ}$ bisects $ AD$ and $ Q\equiv PA\cap O_BO_C$. Our proof is completed then. $ \square$

WLOG $B\ge C$. First, by trivial angle chasing, $\angle{BO_B O_C}=B+2\angle{BAD}$ while $\angle{O_B BC}=90^\circ-\angle{BAD}$, so because $BO_B O_C C$ is cyclic, $\angle{BAD}=90^\circ-\frac{2B+C}{3}$ and $\angle{CAD}=90^\circ-\frac{B+2C}{3}$. We find that $\angle{DAH}=\angle{DAB}-\angle{HAB}=\frac{B-C}{3}$, so it remains to show that $\angle{DAX}=\frac{B-C}{3}$. (*) Note that $\angle{O_B XC}=2\angle{O_B BC}=\frac{4B+2C}{3}$ and $\angle{O_C XB}=2\angle{O_C CB}=\frac{2B+4C}{3}$. However, $\angle{O_B AC}=\angle{O_B AD}+\angle{CAD}=180^\circ-\frac{4B+2C}{3}$, so by symmetry, $ABXO_C$ and $ACXO_B$ are cyclic. Finally, \begin{align*} \angle{DAX}+\angle{DAB} = \angle{XAB} &= \angle{XO_C B} \\ &= \angle{BO_B O_C}-90^\circ = 90^\circ-\frac{B+2C}{3} = \angle{DAB}+\frac{B-C}{3}, \\ \end{align*}so we're done by (*).

Suppose radius of $\odot{BAD}=r_1$ and of $\odot{CAD}=r_2$ and $\angle{DAB}=\theta$ now so $AD=2r_1Sin B=2r_2 Sinc$ also $\angle{O_BDO_C}=\pi-(B+C)=A$ and that implies $O_BDO_C$ is similar to $ABC$ and so $\angle{O_CO_BD}=B,\angle{O_BO_CD}=C$ now also note $B+2\theta+C+\frac{B+\theta}{2}=\pi \implies \theta=\frac{\pi}{2}-\frac {2B+C}{3}$.Now suppose $\angle{XBC}=\alpha$ then $aSin(\alpha)=r_1Sin(\frac {2B+C}{3}-\alpha)$ also $c=2r_1Cos(\frac{B-C}{3})$ , using this we get $\frac{Sin(B-\alpha)}{Sin(C-\alpha)}=\frac{Cos(A-B)}{Cos(A-C)}$ now apply ceva with $AX,BX,CX$ and note $\frac {Sin\angle{XAC}}{Sin\angle{XAB}}=\frac{Cos(A-B)}{Cos(A-C)}=\frac{Cos(\frac{C+2B}{3})}{Cos(\frac{B+2C}{3})}$ and so obviously $\angle{XAB}=\frac{\pi}{2}-\frac{B+2C}{3},\angle{XAC}=\frac{\pi}{2}-\frac{2B+C}{3}$ ,and hence $\angle{XAD}=\frac{B-C}{3}$ and which is indeed $\angle{DAH}$ ,so done!

Without loss of generality $AC > AB$. It is easy to verify via angle chasing that $\angle AO_BB = AO_CC$. Since $O_BO_CCB$ is cyclic, it follows that $A$ is the Miquel point of $O_BO_CCB$. Therefore, $AO_CXB$ is cyclic. Set $x = \angle BAD$, $y = \angle CAD$. Then \[ \angle BO_BO_C = \angle BO_BD + \angle D_OBC = 2x+B \implies \angle BXC = 360 - 4x - 2B \implies \angle BAX = \angle BO_CX = 2x+B-90. \] On the other hand, $\angle BAH = 90 - B$. From here it is easy to derive that $\angle HAD = x+B-90 = \angle XAD$, as desired.

v_Enhance wrote: Without loss of generality $AC > AB$. It is easy to verify via angle chasing that $\angle AO_BB = AO_CC$. Since $O_BO_CCB$ is cyclic, it follows that $A$ is the Miquel point of $O_BO_CCB$. Therefore, $AO_CXB$ is cyclic. Set $x = \angle BAD$, $y = \angle CAD$. Then \[ \angle BO_BO_C = \angle BO_BD + \angle D_OBC = 2x+B \implies \angle BXC = 360 - 4x - 2B \implies \angle BAX = \angle BO_CX = 2x+B-90. \] On the other hand, $\angle BAH = 90 - B$. From here it is easy to derive that $\angle HAD = x+B-90 = \angle XAD$, as desired. How $AO_cXB$ is cyclic? Is it a property..

Yes; see derpy.me/cyclicquad (Fact 12; thanks to Zhero for making this link) Edit: made link clearer; also http://yufeizhao.com/olympiad/cyclic_quad.pdf Linked the other one as well. -- mod

The link isn't working .Paste it in URL Form

Note that $AO_BO_C \sim ABC$ by simple angle chasing, so let $Y = O_BO_C \cap BC$ and $Z = AD \cap O_BO_C$. Notice that by miquel's theorem, we have $Y \in ABO_B$ and $ACO_C$ so since $O_BO_CBC$ are concyclic, we have by radical axis theorem that $BO_B, CO_C, AY$ concur, so $A$ is miquel point of $O_BO_CCB$. Hence, $XA \perp AY$. If we let the midpoint of $O_BO_C$ be $M$ then $\angle DAX = \angle AOM = \angle AYM$ and clearly $\angle HAD = \angle MYD$ (Let $AH \cap BC = E$, then $AYEM$ are concyclic). So, we must show $YM$ bisects $\angle AYE$, or $AM = MQ$ which follows since $M$ is the circumcentre of $AED$.

Sorry for a little glitch in the diagram ($AHG \perp BC, G = AH\cap BC$). Let $BO_B \cap CO_C = Y$. We have $AO_BO_CY, ABCY, BO_BO_CC$ concylic by a little angle chasing, and thus $A$ is the miquel point of cyclic quadrilateral $BO_BO_CC$ (by http://yufeizhao.com/olympiad/cyclic_quad.pdf ) So, we have $XA \perp YA$. Now, let $BC \cap O_CO_B = I$. Then, $AO_BBI$ is cyclic, and $AO_B = BO_B$ implies line $IO_BO_C$ is the bisector of $\angle AIB$. As $AH \perp IB, AD \perp IO_B$ and $AX \perp IA$, $AD$ is the bisector of $\angle HAD$, and thus $\angle DAX = \angle DAH$.

Let $J$ be the circumcenter of $BO_BO_CC$ and $O$ be the circumcenter of $\triangle{ABC}$.Note that $\angle{AO_BO_C}=\angle{AOO_C}=B$ so $A,O_B,O,O_C$ lie on a circle.Also by some angle chasing we obtain that $BO_BO_CC$ is cyclic yeilds that $\angle{ADC}=90+\frac{B-C}{3}$.Next note that $BO_B \cap CO_C=K$ lie on the circumcircle of $ABC$ as well as $AO_BO_C$.Let $r_1,r_2,R$ denote the circumradii of $AO_BO_C,BO_BO_CC$ and $ABC$ respectively. $\angle{BOJ}=\angle{BKO_C}=A$ and $\angle{OBJ}=\angle{OBC}-\angle{JBC}=\angle{BO_CC}-\angle{BKO_C}=\angle{KBO_C}$.Thus $\triangle{OBJ} \sim \triangle{KBO_C} \implies \frac{OJ}{R}=\frac{KO_C}{KB}=\frac{O_BO_C}{BC}=\frac{r_1}{R} \implies OJ=r_1=\frac{R}{2sin\angle{ADC}}=\frac{R}{2cos\frac{(B-C)}{3}}$. Extend $AD$ and $OJ$ so that they meet at $X$.We now get $\frac{AO}{AX}=\frac{OJ}{JX}$ $\Leftrightarrow \frac{OX}{OJ}=\frac{AX}{AO}+1$ $\Leftrightarrow \frac{2cos\frac{(B-C)}{3} \cdot R}{\frac{R}{2cos\frac{(B-C)}{3}}}=\frac{sin(B-C)}{sin\frac{(B-C)}{3}}+1$ $\Leftrightarrow 4cos^2\frac{(B-C)}{3}=4-4sin^2\frac{(B-C)}{3}$ $\Leftrightarrow cos^2\frac{(B-C)}{3}+sin^2\frac{(B-C)}{3}=1$ which is an obvious identity.Therefore $AJ$ bisects $\angle{DAO}$ and consequently $\angle{DAJ}=\angle{DAH}=\frac{B-C}{3}$.

Notice that $\triangle AO_BB$ and $\triangle AO_CC$ are both isoceles. Furthermore, since $O_B$ and $O_C$ are circumcenters, we know that $\measuredangle AO_BB = 2\measuredangle ADB = 2\measuredangle ADC = \measuredangle AO_CC$ where the angles are directed. It follows that $\triangle AO_BB \sim \triangle AO_CC$ with similar orientation. Therefore, $A$ is the center of spiral similarity that sends $\overline{O_BB}$ to $\overline{O_CC}$, so $A$ is the Miquel point of cyclic quadrilateral $BO_BO_CC.$ Hence, it is well-known (and can be proven using harmonic divisions) that $A$ is the inverse of $Y \equiv BO_C \cap CO_B$ with respect to $\odot (BO_BO_CC).$ Then since $C, Y, O_B$ are collinear, under inversion in $\odot (BO_BO_CC)$, it follows that $X, O_B, A, C$ are concyclic. Then because $O_B$ is the circumcenter of $\triangle ABD$ and the points $X, O_B, A, C$ are concyclic, we have \[\measuredangle DAX = \measuredangle O_BAX - \measuredangle O_BAD = \measuredangle O_BCX - \left(90^{\circ} - \measuredangle DBA\right) = \measuredangle O_BCX + \measuredangle DBA - 90^{\circ}.\] Meanwhile, since $AH \perp BC$, we have $\measuredangle HAD = 90^{\circ} - \measuredangle ADB.$ Therefore, \[\measuredangle DAX = \measuredangle HAD \iff \measuredangle O_BCX + \measuredangle DBA + \measuredangle ADB= 180^{\circ} = 0.\] Since the sum of the directed angles in $\triangle ABD$ is equal to $0$, we need only show that $\measuredangle O_BCX = \measuredangle BAD.$ But because $X$ is the circumcenter of $\odot(BO_BO_CC)$, we have \[\measuredangle O_BCX = 90^{\circ} - \measuredangle CBO_B = 90^{\circ} - \measuredangle DBO_B = \measuredangle BAD.\] $\square$

A is Miquel point of BC(OB)(OC). Done

Let $W$ be the circumcircle of triangle $ABC$.and let $W_1$be the circumcircle of triangle $AO_bO_c$.and let $W_2$ be the circumcircle of $BO_bO_cC$.and let $T=W \cap W_1$ By the known lemma we know that lines $AT,O_bO_c,BC$ are concurrent. Let $M$ be their interestion. \[\measuredangle O_bBA = \measuredangle O_cCA\]. So we find that $T=O_bB \cap O_cC$ so we find that $A$ is the Miquel point of $BC(OB)(OC)$ . By the known lemma if $K$ be the circumcenter of $BCO_bO_c$ we find that $MA \perp KA$ By angle casing we find that $MAO_cC$ is cyclice. Now we know that angles$KAD=AMO_c=ACO_c=ABO_b=DAH$

We have that $\angle AO_BO_C = \frac{1}{2}\angle AO_BD = \angle B$ and likewise $\angle AO_CO_B=\angle C$. So $A$ is the center of spiral similarity sending $AO_BO_C$ to $ABC$ and is hence the Miquel point of $BCO_CO_B$. It is well known that $AO_BXC$ is cyclic. So $$\angle XAC = \angle XO_BC = 90-\angle O_BBC = \angle BAD \implies \angle XAD = \angle DAC - \angle BAD.$$Also, $\angle BAH = 90 - \angle B = 2\angle BAD - \angle DAC$, which follows from $\angle BO_BC = \angle O_CCB$. Therefore, $\angle DAH = \angle DAC - \angle BAD$ and the result follows. $\Box$

WLOG, let $AC>AB$. Let $L$ be the midpoint of $\overline{AD},$ and set $Y=\overline{BO_B} \cap \overline{CO_C}, Z=\overline{BC} \cap \overline{O_BO_C}.$ Note that $\triangle AO_BB \sim \triangle AO_CC,$ so $A$ is the spiral center for $\overline{O_BO_C} \mapsto \overline{BC}.$ Consequently, $\overline{YZ} \perp \overline{AX}$. From $\angle ALZ=90^{\circ}$ we conclude that $\overline{AX}$ is tangent to the circle $\odot (AZL).$ Finally, we have $$\angle DAH=\angle DZL=\angle AZL=\angle DAX,$$establishing the result.

BBAI wrote: v_Enhance wrote: Without loss of generality $AC > AB$. It is easy to verify via angle chasing that $\angle AO_BB = AO_CC$. Since $O_BO_CCB$ is cyclic, it follows that $A$ is the Miquel point of $O_BO_CCB$. Therefore, $AO_CXB$ is cyclic. Set $x = \angle BAD$, $y = \angle CAD$. Then \[ \angle BO_BO_C = \angle BO_BD + \angle D_OBC = 2x+B \implies \angle BXC = 360 - 4x - 2B \implies \angle BAX = \angle BO_CX = 2x+B-90. \]On the other hand, $\angle BAH = 90 - B$. From here it is easy to derive that $\angle HAD = x+B-90 = \angle XAD$, as desired. How $AO_cXB$ is cyclic? Is it a property.. Yes it is a property. For cyclic quadrilateral $ABCD$, circles $(AOC)$ and $(BOD)$ concur at the Miquel Point of the quadrilateral, where $O$ is the circumcenter of $(ABCD)$.

anantmudgal09 wrote: WLOG, let $AC>AB$. Let $L$ be the midpoint of $\overline{AD},$ and set $Y=\overline{BO_B} \cap \overline{CO_C}, Z=\overline{BC} \cap \overline{O_BO_C}.$ Note that $\triangle AO_BB \sim \triangle AO_CC,$ so $A$ is the spiral center for $\overline{O_BO_C} \rightarrow \overline{BC}.$ Consequently, $\overline{YZ} \perp \overline{AX}$. From $\angle ALZ=90^{\circ}$ we conclude that $\overline{AX}$ is tangent to the circle $\odot (AZL).$ Finally, we have $$\angle DAH=\angle DZL=\angle AZL=\angle DAX,$$establishing the result. This is a great solution. For anyone stuck on the angle chase(it makes a few jumps) here's a more detailed version. Let $F$ be the foot of the perpendicular from $A$ to $BC$. Observe that since $angle{AFZ}=90$, $F$ lies on $(AZL)$. Now we have $$\angle{DAH}=\angle{LAF}=\angle{LZD}=\angle{DZL}=\angle{AZL}=\angle{DAX}$$.

Really Cute and Easy Problem. USATST 2009 P2 wrote: Let $ ABC$ be an acute triangle. Point $ D$ lies on side $ BC$. Let $ O_B, O_C$ be the circumcenters of triangles $ ABD$ and $ ACD$, respectively. Suppose that the points $ B, C, O_B, O_C$ lies on a circle centered at $ X$. Let $ H$ be the orthocenter of triangle $ ABC$. Prove that $ \angle{DAX} = \angle{DAH}$. Zuming Feng. $\angle AO_BB=2\angle ADB=2(180^\circ-\angle ADC)=\angle AO_CC$, and as $AO_BB$ and $AO_CC$ are isosceles. Hence, there $\exists$ a Spiral Similarity$(\sigma)$ centered at $A$. $\sigma:BO_B\mapsto CO_C\implies\sigma:O_BO_C\mapsto BC$. So, $L=BO_B\cap CO_C=\{\odot(AO_BO_C)\}\cap\{\odot(ABC)\}$. So, by taking Paiwise Radical Axis of $\{\odot(ABC),\odot(BO_BCO_C),\odot(AO_BO_C)\}$ we get that $\{AL,O_BO_C,BC\}$ are concurrent at a point $K$. Now notice that $A$ is the Miquel Point of the cyclic quad $\odot(BO_BO_CC)$. Hence, $\{AX,BO_C,CO_B\}$ are concurrent and also $AX\perp AK$. So, $\angle DAH=\angle DKO_C=\angle O_CKA=\angle DAX\qquad\blacksquare$

[asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ size(8cm); defaultpen(fontsize(10pt)); draw((-1.110459431125061,4.058413889003155)--(-1.62,1.75)--(2.8,1.75)--cycle); /* draw figures */ draw((-1.110459431125061,4.058413889003155)--(-1.62,1.75)); draw((-1.62,1.75)--(2.8,1.75)); draw((2.8,1.75)--(-1.110459431125061,4.058413889003155)); draw(circle((-1.0413773597967102,2.832722400774602), 1.2276367365368883)); draw(circle((1.1686226402032898,3.452813719451886), 2.3581701917399545)); draw(circle((0.59,1.2649171606442673), 2.262610298093205)); draw((1.1686226402032898,3.452813719451886)--(-4.900190463027996,1.75)); draw((-4.900190463027996,1.75)--(-1.62,1.75)); draw((-4.900190463027996,1.75)--(-1.110459431125061,4.058413889003155)); draw((-1.110459431125061,4.058413889003155)--(-0.46275471959342035,1.75)); draw((-1.110459431125061,4.058413889003155)--(0.59,1.2649171606442673)); draw((-1.110459431125061,4.058413889003155)--(-1.110459431125061,1.75)); /* dots and labels */ dot("$A$",(-1.110459431125061,4.058413889003155),NW); dot("$B$",(-1.62,1.75),SW); dot("$C$",(2.8,1.75),E); dot("$D$",(-0.46275471959342035,1.75),S); dot("$O_B$",(-1.0413773597967102,2.832722400774602),dir(150)); dot("$O_C$",(1.1686226402032898,3.452813719451886),N); dot("$T$",(-4.900190463027996,1.75),SW); dot("$M$",(-0.7866070753592407,2.9042069445015777),dir(-30)); dot("$X$",(0.59,1.2649171606442673),NE); dot((-1.110459431125061,1.75)); /* end of picture */ [/asy][/asy] Note $O_BA=O_BB$, $O_CA=O_CC$, and $\measuredangle AO_BB=2\measuredangle ADB=2\measuredangle ADC=2\measuredangle AO_CC$, thus $\triangle AO_BB\sim\triangle AO_CC$ and $A$ is the Miquel point of $BO_BO_CC$. Let $T=\overline{BC}\cap\overline{O_BO_C}$, and let $M$ be the midpoint of $\overline{AD}$. Since $BO_BO_CC$ is cyclic, $A$ is the foot from $X$ to the polar of $\overline{BO_C}\cap\overline{CO_B}$, so $\angle XAT=90^\circ$. Since $\overline{O_BO_C}$ is the perpendicular bisector of $\overline{AD}$, we have $\overline{AD}\perp\overline{TM}$, and since $MA=MD$, we have $\triangle TAD$ isosceles. Hence $\measuredangle DAX=\measuredangle MTA=\measuredangle DTM=\measuredangle HAD$, as desired. (In the final angle chase we use twice the fact that if $a_1\perp b_1$ and $a_2\perp b_2$, then $\measuredangle(a_1,a_2)=\measuredangle(b_1,b_2)$.)

woah cool problem Note that $A$ is the Miquel point of $BO_BO_CC$. Let $M$ be the midpoint of $AD$ and $T$ be the intersection of $O_BO_C$ and $BC$. Let $K$ be the foot of the perpendicular from $A$ to $BC$. Observe that since $M$ is the midpoint of a chord in both $(ABD)$ and $(ACD)$, we have that $\angle TMA = \angle O_BMA = 90$, so $M$ lies on $(AKT)$. Furthermore, note that by a well-known property of Miquel points, $\angle XAT = 90$, so $AX$ is a tangent to $(AMKT)$. Now, we angle chase to finish: $$\angle XAD = \angle XAM = \angle ATM = \angle AKM = \angle KAM = \angle DAH.$$

By the USAMO 2013/6 lemma, $\triangle AO_BO_C\sim\triangle ABC$. Thus $A$ is the Miquel point of $O_BO_CCB$. Let $K=O_BO_C\cap BC$. We need to show that lines $AH,AX$ are isogonal in $\triangle ADK$. Observe that $AX$ is perpendicular to $AK$ by properties of Miquel points and $AH$ is perpendicular to line $DK$. Thus $\angle XAK=90^\circ,\angle HAK=90^\circ-\angle AKD$. We need to show $\angle DAK=90^\circ-\tfrac 12\angle AKD$. But this is immediate by cyclic quad $(KCO_CA)$ from the Miquel configuration and properties of the circumcenter of $\triangle ACD$, thus we are done.

We solve the configuration where $ABD$ is acute and $ACD$ is obtuse; other cases are analogous. Thus angle $B > C$. By angle chasing we get $AO_B O_C, ABC$ are directly similar. Therefore $A$ is the center of the spiral similarity sending $O_B O_C$ to $BC$, thus it is the Miguel Point of $BO_B O_C C$. Let $E = BO_B \cap CO_C$, so $AEO_C O_B, AECB$ are cyclic. Let $\angle ADB = \theta$, so it suffices to show $\angle DAX = 90-\theta$. Remark $AO_B O_C, DO_B O_C$ are congruent. Now angle chasing we get $\angle EBC = B+\theta-90, \angle O_B O_C C = 2\theta - C$. So $\theta = 90.+ \frac{C-B}{3}$. Now this yields $\angle O_B AE = \angle CO_C O_B = 180 - \frac{C}{3}-\frac{2B}{3}$. By Brokard $\angle XAE = 90$, so $\angle O_B AX = 90 - \frac{C}{3}-\frac{2B}{3}$. We can readily find $\angle BAO_B = \frac{B-C}{3}$. So $\angle BAX = 90 - \frac{B}{3}-\frac{2C}{3}=270-B-2\theta$. Hence $\angle DAX = \angle BAX - (180-B-\theta)=90-\theta$. This completes the proof.

Note that $AO_B=BO_B$, $AO_C=CO_C$, and $\measuredangle AO_BB=2\measuredangle ADB=2\measuredangle ADC=\measuredangle AO_CC$. Hence $\triangle AO_BB \sim \triangle AO_CC$, which means that $A$ is the Miquel Point of $BCO_CO_B$. Let $P$ be the intersection of $BC$ and $O_BO_C$ and $Q$ be the foot of the perpendicular from $A$ to $BC$. Since $AD$ is the radical axis of $(ABD)$ and $(ACD)$, $\overline{O_CO_BP}$ is the perpendicular bisector of $AD$. It follows that $\triangle ADP$ is isosceles and $\measuredangle DAP=\measuredangle PDA$. Moreover, $XA$ and $AP$ are perpendicular since the points $ B, C, O_B, O_C$ lies on a circle. Now, $\measuredangle DAX=90-\measuredangle PAD=90-\measuredangle ADP=\measuredangle QAD=\measuredangle HAD$.

Let $E = BO_B \cap CO_C$, $T = BC \cap O_BO_C$, and $F$ be the foot of the $A$-altitude. It's well-known that the Miquel Point of complete quadrilateral $BCO_CO_B$ is $A$. Since $BCO_CO_B$ is cyclic, we know $A \in ET$ and $XA \perp ET$. Claim: $TAD$ is isosceles. Proof. Notice $O_BO_C$, which is the perpendicular bisector of $AD$, passes through $T$. $\square$ Now, $$\angle DAX = 90^{\circ} - \angle TAD = 90^{\circ} - \angle TDA = 90^{\circ} - \angle FDA = \angle DAF = \angle DAH$$as desired. $\blacksquare$ Remarks: I made all the "obvious" observations before getting stuck. In the end, using the fact that $BCO_CO_B$ is cyclic effectively was key. (Instead of angle chasing or Radical Axes, we need to recall the relationship between Brocard's and Miquel Points of cyclic quadrilaterals.)

[asy][asy]import geometry; size(10cm); pen c1=orange+white+white; pen c2=orange+white; pen c3=brown+red+white; pair A,B,C,D,H,O_B,O_C,E,F,X; A=(-2,5); B=(-3,-3); C=(7,-3); D=(0,-3); H=orthocenter(A,B,C); O_B=circumcenter(A,B,D); O_C=circumcenter(A,C,D); E=extension(B,O_B,C,O_C); F=extension(B,C,O_B,O_C); X=circumcenter(B,C,O_B); fill(B--O_B--O_C--C--cycle,c1); draw(B--O_B--O_C--C--cycle,c2); draw(B--F,c2); draw(O_B--F,c2); draw(O_B--E,c2); draw(O_C--E,c2); draw(A--H); draw(A--O_B); draw(A--D); draw(A--X); draw(A--O_C); draw(E--F,dotted); draw(circle(B,C,O_B),c2); draw(circle(A,B,D)); draw(circle(A,C,D)); draw(A--B,c3); draw(A--C,c3); draw(arc(intersectionpoint(bisector(A,C),bisector(A,F)),C,F),dotted); dot(A^^B^^C^^D^^H^^O_B^^O_C^^E^^F^^X); label("$A$",A,NW); label("$B$",B,SW); label("$C$",C,SE); label("$D$",D,S); label("$H$",H,0.6S); label("$O_B$",O_B,SE); label("$O_C$",O_C,NE); label("$E$",E,NE); label("$F$",F,SW); label("$X$",X,NE);[/asy][/asy] WLOG let $AB<AC.$ Let $E=\overline{BO_B}\cap\overline{CO_C}$ and $F=\overline{BC}\cap\overline{O_BO_C}.$ We see that $$\measuredangle AO_BB=2\measuredangle ADB=2\measuredangle ADC=\measuredangle AO_CC,$$$AO_B=BO_B,$ and $AO_C=CO_C,$ so $\triangle AO_BO_C\sim\triangle ABC.$ Hence, the Miquel point of $BO_BO_CC$ is $A,$ and since $BO_BO_CC$ is cyclic, $A$ is the foot from $X$ to $\overline{EF}.$ Notice that \begin{align*}90-\angle DAX&=\angle FAX\\&=180-\angle DFA-\angle ADF\\&=\angle AO_CC+\angle CDA-180\\&=2(180-\angle CDA)+\angle CDA-180\\&=180-\angle CDA\\&=\angle ADB\\&=90-\angle HAD.\end{align*}$\square$

[asy][asy] import olympiad; size(230); /** Coordinates from Geogebra **/ pair A = (-0.36 , 4.12); pair B = (-3 , -2); pair C = (4.64 , -1.82); draw(A--B--C--cycle , green); pair D = (0.27 , -1.92); draw(A--D); draw(circumcircle(A,B,D) , purple); pair O_B = circumcenter(A,B,D); draw(circumcircle(A,C,D) , heavycyan); pair O_C = circumcenter(A,C,D); draw(circumcircle(B , O_C , C) , orange); pair X = circumcenter(B , O_C , C); pair H = orthocenter(A,B,C); draw(circumcircle(A,B,C) , blue); pair O = circumcenter(A,B,C); /** Dots and Labels **/ dot("$A$" , A , N); dot("$B$" , B , SW); dot("$C$" , C , SE); dot("$D$" , D , S); dot("$X$" , X , S); dot("$H$" , H , S); dot("$O$" , O , NE); dot("$O_B$" , O_B , NW); dot("$O_C$" , O_C , NE); [/asy][/asy] Simple angle chasing gives us $AO_BO_C \sim ABC$. Now , let $L = O_BO_C \cap BC$ and $M = AD \cap O_BO_C$. Miquels Theorem gives us $Y \in ABO_B$ and $ACO_C$. Claim: $BO_B , CO_C$ and $AL$ concur. Proof: Since we know that $O_BO_CBC$ are concyclic , we can use the radical axis theorem which completes the proof. This also implies that $A$ is the miquel point of $O_BO_CCB$ $\blacksquare$ Also $XA \perp AL$. Let $G$ be the midpoint of $O_BO_C$ , then we have the following $$\angle DAX = \angle AOM = \angle ALM$$Also $\angle HAD = \angle GLD$. What remains to show is that $LG$ bisects $\angle ALE$ which is true since $G$ is the circumcenter of $\triangle AED$. $\blacksquare$

Let $AB<AC$, $F=BC\cap O_BO_C$. Notice that the point $A$ is the miquel point of the cyclic quadrilateral $BCO_CO_B$. So $XA\perp AF$. Since the line $O_BO_C$ is the perpendicular bisector of $AD$ therefore $\triangle AFD$ is isosceles. So $$\angle XAD=90^\circ-\angle DAF=90^\circ-\angle ADF=\angle DAH$$.

This solution is longer than most other solutions, but hopefully someone will find it helpful. The intuition of simplifying geometry problems works really well here. Angle chase to obtain that we actually want to prove that $\overline{AD}$ and $\overline{AX}$ are isogonal. Take a $\sqrt{bc}$ inversion to get the following problem: I lost the game wrote: Let $D$ be a point on the circumcircle of triangle $ABC$. Let $A_1$ and $A_2$ be the reflections of $A$ over $\overline{BD}$ and $\overline{CD}$, respectively. Prove that if $A_1BCA_2$ has circumcenter $O$, then $\overline{AD}$ and $\overline{AO}$ are isogonal. The crux of the simplifications is this: in fact, we can drop the condition that $A_1BCA_2$ is cyclic, and redefine $O$ as the intersection of the perpendicular bisectors of $\overline{BC}$ and $\overline{A_1A_2}$. Once we see that, the following two claims aren't hard to find. Claim 1: The perpendicular bisector of $\overline{A_1A_2}$ passes through the reflection $A'$ of $A$ over the perpendicular bisector of $\overline{BC}$. Proof: Notice that $DA=DA_1=DA_2$, so $D$ is on the perpendicular bisector of $\overline{A_1A_2}$. Then, proceed with angle chasing. Claim 2: The line isogonal to $\overline{AD}$ and the perpendicular bisector of $\overline{A_1A_2}$ are symmetric about the perpendicular bisector of $\overline{BC}$. Proof: Notice that one line passes through $A$, and the other passes through $A'$. Then, proceed with angle chasing. By claim 2, the line isogonal to $\overline{AD}$ and the perpendicular bisector of $\overline{A_1A_2}$ must intersect on the perpendicular bisector of $\overline{BC}$, so we are done.

Noisss!!! [asy][asy] import graph; size(12cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.5) + fontsize(13); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -0.004, xmax = 0.004, ymin = -0.0055, ymax = 0.0035; /* image dimensions */ draw(arc((-0.0015446764438354107,0.0017601067586798715),0.0005941504392848956,-101.31153334263735,-72.96771295549291)--(-0.0015446764438354107,0.0017601067586798715)--cycle, linewidth(0.75) + blue); draw(arc((-0.0015446764438354107,0.0017601067586798715),0.00047532035142791644,-72.9677129554929,-44.653596847594095)--(-0.0015446764438354107,0.0017601067586798715)--cycle, linewidth(0.75) + blue); draw(arc((-0.0019739622178513794,-0.000386010299822956),0.0005941504392848956,-51.24139176033055,-22.945299768886326)--(-0.0019739622178513794,-0.000386010299822956)--cycle, linewidth(0.75) + blue); /* draw figures */ draw((-0.0015446764438354107,0.0017601067586798715)--(-0.003011434824851063,-0.002313120907518996), linewidth(0.5)); draw((-0.003011434824851063,-0.002313120907518996)--(0.0025781038703707464,-0.002313120907518996), linewidth(0.5)); draw((0.0025781038703707464,-0.002313120907518996)--(-0.0015446764438354107,0.0017601067586798715), linewidth(0.5)); draw(circle((-0.0002166654772401582,-0.0025748828877237124), 0.002807001075993056), linewidth(0.5) + red); draw((xmin, -6.697329579162294*xmin-0.00858510047885425)--(xmax, -6.697329579162294*xmax-0.00858510047885425), linewidth(0.5) + blue); /* line */ draw((-0.0019739622178513794,-0.000386010299822956)--(-0.0015446764438354107,0.0017601067586798715), linewidth(0.5)); draw((-0.0015446764438354107,0.0017601067586798715)--(0.0008208071297595251,0.000031285775362950524), linewidth(0.5)); draw((xmin, -3.264272476241373*xmin-0.003282138041630463)--(xmax, -3.264272476241373*xmax-0.003282138041630463), linewidth(0.5) + blue); /* line */ draw(circle((0.0009335560702811858,0.00014839489907061385), 0.0029562225072598644), linewidth(0.5) + linetype("4 4") + blue); draw(circle((-0.001439392660103263,-0.0005785507672779155), 0.0023410262063539478), linewidth(0.5) + linetype("4 4") + blue); draw((-0.0002166654772401582,-0.0025748828877237124)--(-0.0019739622178513794,-0.000386010299822956), linewidth(0.5)); draw((-0.0019739622178513794,-0.000386010299822956)--(0.0025781038703707464,-0.002313120907518996), linewidth(0.5)); /* dots and labels */ dot((-0.0015446764438354107,0.0017601067586798715),linewidth(3pt) + dotstyle); label("$A$", (-0.0015962278670139036,0.001969223957289919), NE * labelscalefactor); dot((-0.003011434824851063,-0.002313120907518996),linewidth(3pt) + dotstyle); label("$B$", (-0.0034737432551541735,-0.002296776196775628), NE * labelscalefactor); dot((0.0025781038703707464,-0.002313120907518996),linewidth(3pt) + dotstyle); label("$C$", (0.0026341232606945527,-0.0023443082319184196), NE * labelscalefactor); dot((-0.0009364896108516965,-0.002313120907518996),linewidth(3pt) + dotstyle); label("$D$", (-0.001406099726442737,-0.0026057344252037734), NE * labelscalefactor); dot((-0.0019739622178513794,-0.000386010299822956),linewidth(3pt) + dotstyle); label("$O_1$", (-0.0023923894556556638,-0.0002647816944212866), NE * labelscalefactor); dot((0.0008208071297595251,0.000031285775362950524),linewidth(3pt) + dotstyle); label("$O_2$", (0.0007566078725542828,0.0001748896306495358), NE * labelscalefactor); dot((-0.0002166654772401582,-0.0025748828877237124),linewidth(3pt) + dotstyle); label("$X$", (-0.00013461778637306054,-0.0026532664603465647), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); [/asy][/asy] Note that it is enough to prove $\measuredangle XAD=\measuredangle DAH$. Firstly, note that we have a spiral similarity centered at $A$ that maps $\overline{BO_1}\mapsto\overline{CO_2}$. Thus there exists another spiral similarity centered at $A$ that maps $\overline{BC}\mapsto\overline{O_1O_2}$. So $A$ is the miquel point of the quadrilateral $BCO_2O_1$. Moreover, as we are given that $BCO_1O_2$ is cyclic, we thus get the $AX$ is the angle-bisector of $\angle O_1AC$ and $\angle BAO_2$. Moreover, due to this same miquel point config, we also have that $AO_2XB$ and $AO_1XC$ are also cyclic. Now the condition is equivalent to the following, \begin{align*} &\measuredangle XAD=\measuredangle DAH\\ &\iff \measuredangle XAD+\measuredangle HAB=\measuredangle DAH+\measuredangle HAB\\ &\iff \measuredangle XAD+(\measuredangle CBA+90^\circ)=\measuredangle DAB\\ &\iff \measuredangle XAD+\measuredangle DAO_1=\measuredangle DAB\\ &\iff \measuredangle XAO_1=\measuredangle CAX .\end{align*} Thus it suffices to prove that $AD$ and $AX$ are pairs of isogonals w.r.t. $\angle BAC$. Now comes the final angle chase. (I felt that this was really messy IDK why lol (maybe due to my nubness )) Firstly, we have $\measuredangle XO_1C=90^\circ-CBO_1\implies \measuredangle CO_1X=90^\circ+\measuredangle CBO_1$ and we also have $\measuredangle O_1BD=90^\circ-\measuredangle DAB\implies \measuredangle DAB=90^\circ-\measuredangle O_BD$. Now we just follow through the rest, \begin{align*} \measuredangle CAX&=\measuredangle CO_1X\\ &=\measuredangle CBO_1+90^\circ\\ &=\measuredangle CBA-\measuredangle O_1BA+90^\circ\\ &=\measuredangle DBO_1+90^\circ\\ &=90^\circ-\measuredangle O_1BD\\ &=\measuredangle DAB ,\end{align*} and we are done!

Let $P = O_BO_C \cap BC$, which lies on the perpendicular bisector of $AD$. Then $A$ is the Miquel point of cyclic quadrilateral $O_BO_CCB$, so \[\measuredangle DAH = \measuredangle O_BPD = \measuredangle APO_B = \measuredangle XAD. \quad \blacksquare\]

$A$ is the Miquel Point of $BCO_CO_B$ hence lies on $(BXO_C)$ and $(CXO_B)$. Angle chase to the finish.

We first make the following claim: Claim 1: There exists a spiral similarity at $A$ sending $O_BB \stackrel{A}{\mapsto} O_CC$, so that $\triangle AO_BB \sim \triangle AO_CC$. Proof: We have $\angle AO_BB = 2 \angle ADB$, and \begin{align*} \angle AO_CC &= 360^{\circ} - 2 \angle ADC \\ &= 360^{\circ} - 2(180^{\circ} - \angle ADB) \\ &= 2 \angle ADB. \phantom{c} \square \end{align*} Now let $Y = BO_B \cap CO_C$. Then we make another claim: Claim 2: $AYCB$ and $AYO_CO_B$ are cyclic. Proof: We have \begin{align*} \angle BYC &= 180^{\circ} - (\angle YBC + \angle YCB) \\ &= 180^{\circ} - (\angle B - \cancel{\angle ABO_B} + \angle C + \cancel{\angle O_CCA}) \\ &= \angle A. \end{align*}Now by our first claim, $\angle O_BA_C = \angle A$, so that $AYO_CO_B$ is cyclic too. $\square$ Let $Z = O_BO_C \cap BC$. Then as $X$ is the circumcenter of cyclic $BO_BO_CC$ and $A$ is the Miquel point of quadrilateral $BO_BO_CC$, we find that $XA \perp ZA = YA$. Now observe $ZO_B$ bisects $\angle AO_BB,$ as $AO_BBZ$ is cyclic. From the following $ZB \perp AH, AD \perp ZO_B,$ and $AX \cap ZA$, we find that \[\angle HAD = 90^{\circ} - \angle ADB = \frac{1}{2}(180^{\circ} - \angle AO_BB) = \frac{1}{2} \angle AZB,\]and \[\angle DAX = \angle AZO_B = \frac{1}{2} \angle AZB,\]so that $AD$ bisects $\angle HAX$, as desired. $\blacksquare$