Consider the following Lemma. $\angle A_1OC_1=180^{\circ}-2\angle ABC$. (Proof) Let $L$ be the midpoint of $BH$; $A'$ and $C'$ be points on $BC$ and $BA$, respectively, such that $A'A_1$ and $C'C_1$ are altitudes in $\triangle BA'C'$. Notice that $\triangle BA'C' \sim \triangle BC_1A_1$ and the ratio of similarity is $\cos A$. Also, $BO \perp A'C'$ and $\tfrac{BL}{BO}=\cos A$; so $O$ is the foot of the altitude in $\triangle BA'C'$. Consequently, $\angle A_1OC_1=180^{\circ}-2\angle ABC$. $\blacksquare$ Note that $A_1C_1 \parallel AC$. Let $O_1$ be the circumcenter of triangle $A_1BC_1$; then $B, O_1, O$ are collinear. Also, $\angle A_1O_1C_1+\angle A_1OC_1=180^{\circ}$ so $A_1, O, C_1, O$ are concyclic. Hence, we conclude that $OB$ bisects angle $A_1OC_1$. $\blacksquare$

Alternatively, Apply inversion at $B$ with power $\sqrt{BA_1 \cdot BC}$, then $\{A, A_1\}, \{C, C_1\}$ are swapped. Hence, $\{H, O\}$ are swapped. Consequently, $\measuredangle A_1OB=\measuredangle C_1OB$ since $\measuredangle BAH=\measuredangle BCH$, so we are done. $\blacksquare$

This problem is taken from Crux.Look at Yufei Zhao's Similarity Handout and you will find this question.

I don't think that's much of an issue. When I was an 8-th grader, I could barely draw the figure, if not rage quit, a problem like this

$B'$ is the diameter of $(O)$ . $BH,CH$ cuts $(O)$ at $X,Y$ ,$B'X,B'Y$ cuts $BA,BC$ at $M,N$ . Then $\angle B'XC = \angle B'AC = \angle XHM$ so $HM \parallel BC \equiv HN$ . But $B'B$ is the bisector of $\angle YB'X$ so $OB$ is the bisector of $A_1OC_1$

http://yufeizhao.com/olympiad/similarity.pdf Problem 7

Another way: Let $M,N$ be points on $BC$ so that $OM\parallel AB, ON\parallel BC$. Since $BH=2OX, X$ midpoint of $AC$, $\triangle MON\equiv\triangle A_1BC_1$ and easily $\triangle AOM\equiv\triangle OBA_1,\triangle OCN\equiv\triangle BC_1O$. With $\angle ACO=\angle OCA$, done. Best regards, sunken rock

anantmudgal09 wrote: Consider the following Lemma. $\angle A_1OC_1=180^{\circ}-2\angle ABC$. (Proof) Let $L$ be the midpoint of $BH$; $A'$ and $C'$ be points on $BC$ and $BA$, respectively, such that $A'A_1$ and $C'C_1$ are altitudes in $\triangle BA'C'$. Notice that $\triangle BA'C' \sim \triangle BC_1A_1$ and the ratio of similarity is $\cos A$. Also, $BO \perp A'C'$ and $\tfrac{BL}{BO}=\cos A$; so $O$ is the foot of the altitude in $\triangle BA'C'$. Consequently, $\angle A_1OC_1=180^{\circ}-2\angle ABC$. $\blacksquare$ After this lemma, we observe that $H'$ is the orthocenter of $\triangle A'BC'$ also the incenter of $\triangle A_1 O C_1$ we get immediately the result.

This solution seems different from the above ones: It is well known that the homothety at the nine point center $N$ with ratio $-1$ takes the line $AC$ to the line $A_1C_1$. Notice that this homothety takes $A$ to the reflection of $O$ in $BC$. Hence the reflection of $OC_1$ about $BC$ is $A_1C_1$. Similarly, the reflection of $OA_1$ about $BA$ is the line $A_1C_1$. So, $B$ is the excenter of triangle $OA_1C_1$. Hence the conclusion. $\blacksquare$.

It can also be proved by proving that B is the excenter of triangle A1OC1

Or a surprisingly easy trig bash: Let $OB$ meet $A_1C_1$ in $D$. Note that triangle $BA_1C_1$ is obtained from triangle $BAC$ by a homothety with scale factor $cosB/2sinAsinC$. Sine rule in triangle $BA_1C_1$ gives that $A_1D/C_1D = sin2C/sin2A$ To get $OA_1/OC_1 = sin2C/sin2A$ use the cosine law about angles $A_1BO$ and $C_1BO$.

Let $B'$ be $B-$antipode and $C',A'$ be reflection of $H$ over $AB,AC$. Apply Pascal on $C'B'A'ABC$ $\implies$ $BO$ bisects $\angle XB'Y$. Let $B'C' \cap AB=X$ and $B'A' \cap BC=Y$ $\implies$ $X-Y-H$. Since, $\angle XHC'$ $=$ $\angle XC'H$ $=$ $90^{\circ}-A$ $=$ $\angle C'CA$ $\implies$ $XH$ $||$ $AC$ $\implies$ $XY||AC$. Now take a homothety at $B$ with ratio $\tfrac12$

Dear Mathlinkers, http://jl.ayme.pagesperso-orange.fr/Docs/Orthique%20encyclopedie%204.pdf p. 10-17. Sincerely Jean-Louis

wait can't u just reflect H over AB and CB to get F and D? then you get that $A_1$ is the circumcenter of FBH, and then it's a really easy angle chase.

Sharygin Finals 2017, Problem 8.2 wrote: Let $H$ and $O$ be the orthocenter and circumcenter of an acute-angled triangle $ABC$, respectively. The perpendicular bisector of $BH$ meets $AB$ and $BC$ at points $A_1$ and $C_1$, respectively. Prove that $OB$ bisects the angle $A_1OC_1$. Solution. Note that $A_1C_1\parallel AC$, $\angle ADH = 2 \angle C = \angle AOB$, also $C_1A=C_1H$, and $OB=OC\implies \triangle BC_1H\sim BOA$. Hence, there exists a spiral similarity centered at $B$, which sends $C_1\mapsto O$, $H\mapsto A\implies \triangle BC_1O\sim \triangle BHA$. Similarly, $\triangle BA_1O\sim \triangle BHC$. $$\angle BOC_1=\angle BAH=90^\circ -\angle B=\angle BCH=\angle BOA_1.\blacksquare$$

Notice that $\angle A_1HC_1+\angle AHC=180^\circ$ therefore $H$ has its isogonal conjugate wrt quadrilateral $ACC_1A_1$, and it's isogonal conjugate is $O$. Let $O_1$ be the circumcenter of $(BA_1C_1)$ which lies on $BO$. Notice $180^\circ=\angle AOC+\angle A_1OC_1=\angle A_1O_1C_1+\angle A_1OC_1$. So $A_1OC_1O_1$ is cyclic. Since $A_1O_1=C_1O_1$ therefore $OO_1$ bisects $\angle A_1OC_1$.

This took me way longer than it should have. Let $M$ denote the midpoint of $BH$, and $\measuredangle$ denote any angle modulo $\pi$. Observe that as $A_1M \perp BH$, we have that $\triangle BA_1H$ (and similarly $\triangle BC_1H$) are separately isosceles. We make the following claim: Claim: $\triangle BA_1H \sim \triangle BOC$ and $\triangle BC_1H \sim \triangle BOA$. Proof: We have that \[\measuredangle OCB = \measuredangle CBO = \measuredangle HBA_1 = \measuredangle A_1HB. \]A similar process ensues for the second similarlity. $\square$ At present, there exists a spiral similarity at $B$ sending $C_1 \mapsto O, H \mapsto A$ and $A_1 \mapsto O, H \mapsto C$. Therefore we have another spiral similarity at $B$ so that $\triangle BC_1O \sim \triangle BHA$ and $\triangle BA_1O \sim \triangle BHC$. Hence we have that \begin{align*} \measuredangle A_1OB &= \measuredangle HCB \\ &= \measuredangle BAH \\ &= \measuredangle BOC_1, \end{align*}as desired. $\blacksquare$