I did this with complex numbers. Let $AH,BH,CH$ intersect the circumcircle of $\triangle ABC$ again at $A'B'C'$. $X,Y,Z$ are the reflections of $A_1,B_1,C_1$ in lines $BC,CA,AB$. Let $P$ be the orthocentre of $\triangle H_1H_2H_3$. It suffices to prove $A_1P\perp B_1C_1$ as the other two relations will follow similarly. Now we do some really easy Euclidean geometry - it is well known that the circle $BCH$ is a reflection of circle $ABC$ in $BC$. So $H_1,H_2,H_3,X,Y,Z$ all lie on the circumcircle of $\triangle ABC$. $XA'=A_1H$ etc. so $XA'=YB'=ZC'$. The foot of the perpendicualar $A$ to $BC$ is $\frac{1}{2}(a+b+c - \frac{bc}{a})$ and $BC$ bisects $HA'$ so $a' =-\frac{bc}{a}$. (We used the well-known fact that $h=a+b+c$) We get similar expressions for $b',c'$. $XA'=YB'$ so $XY||A'B'$ or $XB'||YA'$ So $xy=a'b'$ or $xb'=ya'$ so $y=\frac{c^2}{x}$ or $\frac{xa^2}{b^2}$. Similarly $z=\frac{b^2}{x}$ or $\frac{xa^2}{c^2}$. So $b_1 = a+c-\frac{ac}{y}$ $b_1=a+c-\frac{ax}{c}$ or $a+c-\frac{b^2c}{ax}$ and $c_1=a+b-\frac{ab}{z}$. $c_1=a+b-\frac{ax}{b}$ or $a+b-\frac{bc^2}{ax}$. also $a_1=b+c-\frac{bc}{x}$. $H_1$ is the point at which the perpendicular from $X$ to $BC$ intersects the circle $ABC$ again. So $h_1=-\frac{bc}{x}$. Similarly we get formulas for $h_2,h_3$. $p=h_1+h_2+h_3=-\frac{bc}{x}-\frac{ca}{y}-\frac{ab}{z}$. $\frac{a_1-p}{b_1-c_1}=\frac{b+c+\frac{ca}{y}+\frac{ab}{z}}{b_1-c_1}$. Case 1: $y=\frac{c^2}{x},z=\frac{b^2}{x}$. $\frac{a_1-p}{b_1-c_1}=\frac{b+c+\frac{ax}{c}+\frac{ax}{b}}{c-b+\frac{ax}{b}-\frac{ax}{c}} = \frac{(b+c)(bc+ax)}{(c-b)(bc+ax)} = \frac{b+c}{c-b}$ Upon conjugation, this changes sign, so $A_1P\perp B_1C_1$. Case 2: $y=\frac{c^2}{x},z=\frac{xa^2}{c^2}$. $\frac{a_1-p}{b_1-c_1}=\frac{b+c+\frac{ax}{c}+\frac{bc^2}{ax}}{c-b-\frac{ax}{c}+\frac{bc^2}{ax}}$ $=\frac{a^2x^2+xac(b+c)+bc^3}{-a^2x^2+xac(c-b)+bc^3}$ $=\frac{(axc+bc^2)(\frac{ax}{c}+c)}{(axc+bc^2)(c-\frac{ax}{c})}$ $=\frac{ax+c^2}{c^2-ax}$. Upon conjugation this changes sign so $A_1P\perp B_1C_1$. Case 3: $y=\frac{xa^2}{b^2},z=\frac{b^2}{x}$ Same as Case2, just that $b,c$ are switched around and you have a change of sign. Case 4: $y=\frac{xa^2}{b^2},z=\frac{xa^2}{c^2}$. $\frac{a_1-p}{b_1-c_1}=\frac{b+c+\frac{cb^2}{ax}+\frac{bc^2}{ax}}{c-b-\frac{b^2c}{ax}+\frac{bc^2}{ax}}=\frac{ax(b+c)+bc(b+c)}{ax(c-b)-bc(b-c)} =\frac{b+c}{c-b}$. Upon conjugation it changes sign so we're done.

Mladenov wrote: Let $H$ be the orthocenter of $\triangle ABC$. The points $A_{1} \not= A$, $B_{1} \not= B$ and $C_{1} \not= C$ lie, respectively, on the circumcircles of $\triangle BCH$, $\triangle CAH$ and $\triangle ABH$ and satisfy $A_{1}H=B_{1}H=C_{1}H$. Denote by $H_{1}$, $H_{2}$ and $H_{3}$ the orthocenters of $\triangle A_{1}BC$, $\triangle B_{1}CA$ and $\triangle C_{1}AB$, respectively. Prove that $\triangle A_{1}B_{1}C_{1}$ and $\triangle H_{1}H_{2}H_{3}$ have the same orthocenter. Let $(O, R)$ be circumcircle of $\triangle ABC$ and $(O_1), (O_2), (O_3)$ reflections of $(O)$ in $BC, CA, AB.$ These 3 circles concur at orthocenter $H$ of $\triangle ABC$ $\implies$ $(O_1) \equiv \odot(HBC),$ $(O_2) \equiv \odot(HCA),$ $(O_3) \equiv \odot(HAB).$ Let $(H)$ be circle with center $H$ and arbitrary radius less than $2R$, intersecting $(O_1)$ at 2 points $A_1, A_2,$ $(O_2)$ at 2 points $B_1, B_2$ and $(O_3)$ at 2 points $C_1, C_2$ $\implies$ $\exists$ 2 points $A_1, A_2 \in (O_1),$ 2 points $B_1, B_2 \in (O_2)$ and 2 points $C_1, C_2 \in (O_3),$ such that $A_1H = A_2H = B_1H = B_2H = C_1H = C_2H.$ Assume that points $H, A_1, A_2$ lie in $(O_1)$ counterclockwise, points $H, B_1, B_2$ lie in $(O_2)$ counterclockwise and points $H, C_1, C_2$ lie in $(O_3)$ counterclockwise. Case 1: Choose points $\{A_1, B_1, C_1\}.$ Proof for points $\{A_2, B_2, C_2\}$ would be exactly the same. Perpendiculars to $BC, CA, AB$ through $A_1, B_1, C_1$ cut $(O)$ at $H_1, H_2, H_3,$ such that $H_1, H_2, H_3$ are not reflections of $A_1, B_1, C_1$ in $BC, CA, AB$ $\implies$ $H_1, H_2, H_3$ are orthocenters of $\triangle A_1BC, \triangle B_1CA, \triangle C_1AB.$ $\overline{H_1A} = \overline{A_1H}, \overline{H_2B} = \overline{B_1H}, \overline{H_3C} = \overline{C_1H}$ $\implies$ $\measuredangle H_1OA = \measuredangle H_2OB = \measuredangle H_3OC$ $\implies$ $\triangle H_1H_2H_3 \cong \triangle ABC$ with the same circumcircle $(O)$ are congruent. $H, B_1 \in (O_2),$ $H, C_1 \in (O_3),$ $(O_2) \cong (O_3)$ and $HB_1 = HC_1$ $\implies$ $\measuredangle HAB_1 = \measuredangle HAC_1$ $\implies$ points $A, B_1, C_1$ are collinear. Likewise, points $B, C_1, A_1$ are collinear and points $C, A_1, B_1$ are collinear $\implies$ $\measuredangle C_1A_1B_1 = \measuredangle CAB,$ $\measuredangle A_1B_1C_1 = \measuredangle ABC$ and $\measuredangle B_1C_1A_1 = \measuredangle BCA$ $\implies$ $\triangle A_1B_1C_1 \sim \triangle ABC$ are similar. Let $K$ be orthocenter of $\triangle H_1H_2H_3$ $\implies$ $H_1A_1 = AH = H_1K,$ $H_2B_1 = BH = H_2K,$ $H_3C_1 = CH = H_3K$ $\implies$ $\triangle KH_1A_1, \triangle KH_2B_1, \triangle KH_3C_1$ are $H_1-, H_2-, H_3-$ isosceles. Let $V \equiv A_1H_1 \cap B_1H_2.$ Since $A_1H_1 \perp BC$ and $B_1H_2 \perp CA$ $\implies$ $\measuredangle H_2VH_1 = 180^\circ - \measuredangle BCA = \measuredangle H_2KH_1$ $\implies$ $KH_1H_2V$ is cyclic $\implies$ $\measuredangle KH_2B_1 \equiv \measuredangle KH_2V = \measuredangle KH_1V \equiv KH_1A_1$ $\implies$ isosceles $\triangle KH_1A_1 \sim \triangle KH_2B_1$ are similar. Likewise, isosceles $\triangle KH_2B_1 \sim \triangle KH_3C_1$ and isosceles $\triangle KH_3C_1 \sim \triangle KH_1A_1$ are similar $\implies$ $\triangle A_1B_1C_1 \sim \triangle H_1H_2H_3$ are spirally similar with similarity center $K$ (their common orthocenter), similarity coefficient $\frac{KA_1}{KH_1} = \frac{KB_1}{KH_2} = \frac{KC_1}{KH_3}$ and rotation angle $\measuredangle A_1KH_1 = \measuredangle B_1KH_2 = \measuredangle C_1KH_3$. Case 2: Choose points $\{A_1, B_1, C_2\}.$ Proofs for the remaining point combinantions $\{A_1, B_2, C_1\},$ $\{A_2, B_1, C_1\},$ $\{A_2, B_2, C_1\},$ $\{A_2, B_1, C_2\},$ $\{A_1, B_2, C_2\}$ would be exactly the same. Perpendiculars to $BC, CA, AB$ through $A_1, B_1, C_2$ cut $(O)$ at $H_1, H_2, H_3,$ such that $H_1, H_2, H_3$ are not reflections of $A_1, B_1, C_2$ in $BC, CA, AB$ $\implies$ $H_1, H_2, H_3$ are orthocenters of $\triangle A_1BC, \triangle B_1CA, \triangle C_2AB.$ $H_1A = A_1H = C_2H = H_3C$ and $H_2B = B_1H = C_2H = H_3C$ $\implies$ $AH_1H_3C$ asnd $BH_2H_3C$ are both isosceles trapezoids $\implies$ $H_1H_3 \parallel AC, H_2H_3 \parallel BC.$ Congruent circles $(O_2) \cong (O_3)$ are symmetrical WRT their radical axis $AH$. Since $B_1H = C_2H$, points $B_1 \in (O_2)$ and $C_2 \in (O_3)$ are also symmetrical WRT $AH$ $\implies$ $B_1C_2 \perp AH$ $\implies$ $B_1C_2 \parallel BC$ and likewise, $A_1C_2 \parallel AC$. It follows that $H_2H_3 \parallel BC \parallel B_1C_2$ and $H_1H_3 \parallel AC \parallel A_1C_2$. Since $A_1H_1 \perp BC$ and $B_1H_2 \perp AC$ $\implies$ $A_1H_1, B_1H_2$ is a pair of common altitudes of $\triangle A_1B_1C_2, \triangle H_1H_2H_3,$ intersecting at their common orthocenter $K$.

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