After some angle chasing in main problem , you obtain $\measuredangle (\ell _b , \ell _c) = \widehat{A}$ and so $\ell _b \cap \ell _c$ , $A$ , $B''$ , $C''$ are concylic .To finish the solution , you need to prove that orthocenter also lies on latter circle , which is a particular case of following theorem : Own theorem (of course , until someone link to some reference ) Let $P$ an arbitrary point . $A' , B' , C'$ varies on sides $BC , CA , AB$ of triangle $\triangle ABC$ such that $(AB'C') , (BC'A') , (CA'B')$ all pass through $P$ . Let $A'' , B'' , C''$ be second intersection of $(A'B'C')$ with sides of $\triangle ABC$ . Then $(AB''C'') , (BC''A'') , (CA''B'')$ pass through $P'$ which is isogonal conjugate of $P$ WRT $\triangle ABC$.

We have that $\angle OA'B'=\angle OCB'=\angle OCA=90^{\circ}-B$ and similarly $\angle OA'C'=90^{\circ}-C$. Therefore $\angle B'A'C'=90^{\circ}-B+90^{\circ}-C=A$. Symmetric arguments yield $\angle A'B'C'=B$ and $\angle B'C'A'=C$. Suppose $(A'B'C')$ intersects $BC,CA,AB$ at $D,E,F$ such that the 6 points $A',B'C',D,E,F$ are all distinct. Now $\angle C'DA'=\angle C'B'A'=B=\angle C'BD$ so triangle $C'BD$ is isosceles. Thus $C'B=C'D$. Similarly $B'D=B'C$. Now, let $P$ be the reflection of $A$ in the diameter of $(AB'C')$ through $O$. We shall prove that $P$ is the reflection of $D$ in $B'C'$. Now $OA=OP$ so $P$ lies on $(ABC)$. Then $\angle AB'P=\angle AOP=2\angle ACP$ which implies $\angle CB'P=180^{\circ}-2\angle ACP$ hence $\angle B'CP=\angle B'PC$. So triangle $B'CP$ is isosceles and $B'P=BC$. A similar argument shows that $C'P=C'B$. Therefore $P$ is the reflection of $D$ in $B'C'$ since $B'P=BC=BD$ and $C'P=C'B=CD$. It therefore follows that the radical axis of $(B',B'C)$ and $(C',C'B)$ is $PD$ since $P$ and $D$ are the intersection of these two circles. So $\ell_a=PD$. Note that $PD\perp B'C'$ since $P$ is the reflection of $D$ through $B'C'$. So we can now disregard $P$ and consider $\ell_a$ as the line through $D$ perpendicular to $B'C'$. Similarly, $\ell_b$ is the line through $E$ perpendicular to $C'A'$ and $\ell_c$ is the line through $F$ perpendicular to $A'B'$. Let $\ell_b$ and $\ell_c$ intersect at $X$, $\ell_c$ and $\ell_a$ intersect at $Y$ and $\ell_a$ and $\ell_b$ intersect at $Z$. So we aim to prove the orthocentre of triangle $ABC$ is the same as the orthocentre of triangle $XYZ$. Suppose $\ell_a,\ell_b,\ell_c$ meet $B'C',C'A',A'B'$ at $L,M,N$ respectively. Now, because $XM\perp C'A'$ and $XN\perp A'B'$, the points $X,M,N$ and $A'$ lie on a circle with diameter $XA'$. It follows that $\angle YZX=\angle M'A'N'=\angle C'A'B'=A$. Repeat these arguments for $\angle XYZ$ and $\angle YZX$ to find $\triangle ABC\sim\triangle A'B'C'\sim\triangle XYZ$. Consider the orthocentre $H_1$ of $\triangle AB'C'$. Since $\angle C'H_1B'=180^{\circ}-A$ (easy angle-chasing) we have that $\angle C'H_1B'+\angle C'A'B'=180^{\circ}$ so $H_1$ lies on $(A'B'C')$ (on the arc $B'C'$ not containing $A'$). Now note that $BF=BA$ (we proved earlier $C'B=C'D$ and $B'C=B'D$, this is the analogous result). So the perpendicular to $AF$ through $B$ bisects the segment $AF$ since triangle $B'AF$ is isosceles. Therefore $BH_1$ bisects $AF$, i.e. $BH_1$ is the perpendicular bisector of $AF$. Similarly, $CH_1$ is the perpendicular bisector of $AE$ and it follows $H_1$ is the circumcentre of triangle $AEF$. Note that since $\angle FXE=180^{\circ}-\angle YZX=180^{\circ}-A=180^{\circ}-\angle EAF$, we have that $AEXF$ is cyclic, and the centre of this circle is $H_1$. Similarly define $H_2$ and $H_3$. Next, note by Miquel's theorem the circles $(AEF),(BFD)$ and $(CDE)$ concur. Let the common point be $H'$. Now it is well known that the if $T_1T_2T_3$ is a triangle with incentre $I$ and $S_1$ is the midpoint of the arc $T_2T_3$ on $(T_1T_2T_3)$ not containing $T_1$ then $(S_1,S_1T_2)$ passes through $T_2,T_3$ and $I$ (just simple angle-chasing). Note that $H_1$, since it is the circumcentre of triangle $AEF$, lies on the perpendicular bisector of $EF$, but it also lies on $(A'B'C'DEF)$ so $H_1$ must be the midpoint of the arc $EF$ not containing $D$. Therefore, the incentre of triangle $DEF$ lies on $(H_1,H_1E)=(AEF)$. Similar arguments imply that the incentre of triangle $DEF$ lies on $(BFD)$ and $(CDE)$. This forces $H'$ to be the incentre. Therefore, $DH'$ bisects $\angle EFD$, and so $D,H'$ and $H_1$ are collinear. Similarly $E,H'$ and $H_2$ are collinear; $F,H'$ and $H_3$ are collinear. Thus, $\angle H_1H'E=180^{\circ}-DH'E=\angle ECD=C$. Therefore since $H_1H'=H_1E$, we have $\angle H_1EH'=\angle H_1H'E=C\implies\angle EH_1H'=180^{\circ}-2C$ so $\angle DFH'=\angle EFH'=\frac{1}{2}\angle DFE=\frac{1}{2}\angle DH_1E=90^{\circ}-C$. Therefore $\angle H'BD=\angle DFH'=90^{\circ}-C$ which implies $\angle H'BC=90^{\circ}-C$. It follows that $BH'\perp CA$. Similarly $CH'\perp AB$ and $AH'\perp BC$ so $H'$ is the orthocentre $H$ of triangle $ABC$. It remains to prove that $H=H'$ is the orthocentre of triangle $XYZ$. Now $\angle HCD=90^{\circ}-B\implies\angle HZD=90^{\circ}-B\implies \angle HZY=90^{\circ}-B$. Since $\angle ZYX=B$, we can see that $ZH\perp XY$. Similarly $YH\perp XY$ and $XH\perp YZ$. So $H$ is the orthocentre of triangle $XYZ$ and this proves the result. Remark: An interesting result is that if $T$ is the circumcentre of triangle $XYZ$ then $OT=OH$ and the circumcentre of $(OTH)$ coincides with the centre of $(A'B'C'DEF)$.

mahanmath wrote: Own theorem (of course , until someone link to some reference ) Let $P$ an arbitrary point . $A' , B' , C'$ varies on sides $BC , CA , AB$ of triangle $\triangle ABC$ such that $(AB'C') , (BC'A') , (CA'B')$ all pass through $P$ . Let $A'' , B'' , C''$ be second intersection of $(A'B'C')$ with sides of $\triangle ABC$ . Then $(AB''C'') , (BC''A'') , (CA''B'')$ pass through $P'$ which is isogonal conjugate of $P$ WRT $\triangle ABC$. Do you have a proof of this theorem? I've seen the following result in The Algebra of Geometry by Christopher J Bradley: Let $P$ be an arbitrary point not on the sides of triangle $ABC$. Suppose that $Q$ is the isogonal conjugate of $P$ with respect to triangle $ABC$. Let $LMN$ be the pedal triangle of $P$ and $DEF$ be the pedal triangle of $Q$. Then $L,M,N,D,E,F$ lie on a circle. Of course, this is weaker than your result However, Bradley also proves that the centre of this circle is the midpoint of $PQ$. For your version, the centre of the circle lies on the perpendicular bisector of $PP'$ (see my remark at the end of my previous post).

WakeUp wrote: Do you have a proof of this theorem? I've seen the following result in The Algebra of Geometry by Christopher J Bradley: Let $P$ be an arbitrary point not on the sides of triangle $ABC$. Suppose that $Q$ is the isogonal conjugate of $P$ with respect to triangle $ABC$. Let $LMN$ be the pedal triangle of $P$ and $DEF$ be the pedal triangle of $Q$. Then $L,M,N,D,E,F$ lie on a circle. Yes , I have a quick proof which use the celebrated theorem that you mentioned . First prove that $(AB''C'')$ pass though a fixed point , and then use above theorem to prove that this point is exactly $P'$

Nice problem!

very nice problem here is my solution in steps i will write all steps that are not angle chasing let $A'',B'',C''$ be the intersections of $\odot A'B'C'$ and $BC,CA,AB$ resp. first angle chase to $\triangle A'B'C'\sim \triangle ABC$ than $\angle B'A''C=\angle A'B'C'=\angle B'CA''$ so $B'C=B'A''$ simirarly $C'A''=C'B$ now $l_{a}$ is the line through $A''$ perp to $B'C'$. we get the same thing for $B''$ and $C''$ similarly. now let $l_{a}\cap l_{b}=Z$ similarly define $X,Y$ angle chase to $\triangle XYZ\sim \triangle ABC$. now let $H$ be the incenter of $B''C''A''$ first angle chase to $HA''CB''$ being cyclic than to $CH\perp AB$ and similarly for $BH,AH$ to get that $H$ is orthocenter of $\triangle ABC$ now angle chase to $ZB''CA''$ being cyclic and with $B''CA''H$ cyclic we get $ZHA''C$ cyclic and finally angle chase to $ZH\perp l_{c}$ and simirarly we get for $XH,YH$ to finally get that $H$ is the orthocenter of $XYZ$ as well.

goodar2006 wrote: Let $O$ be the circumcenter of the acute triangle $ABC$. Suppose points $A',B'$ and $C'$ are on sides $BC,CA$ and $AB$ such that circumcircles of triangles $AB'C',BC'A'$ and $CA'B'$ pass through $O$. Let $\ell_a$ be the radical axis of the circle with center $B'$ and radius $B'C$ and circle with center $C'$ and radius $C'B$. Define $\ell_b$ and $\ell_c$ similarly. Prove that lines $\ell_a,\ell_b$ and $\ell_c$ form a triangle such that it's orthocenter coincides with orthocenter of triangle $ABC$. Proposed by Mehdi E'tesami Fard Nice problem! Let $H$ be the orthocenter of $\triangle ABC$. Note that $\triangle A'B'C' \sim \triangle ABC$. Let $\odot(A'B'C')$ meet $\overline{BC}, \overline{CA}, \overline{AB}$ at $D,E,F$ respectively. Then $\measuredangle BDC'=-\measuredangle A'B'C'=-\measuredangle ABC$ hence $C'B=C'D$. Likewise $B'D=B'C$ hence $D \in \ell_a$. Let $\odot(AB'C')$ meet $\odot(ABC)$ again at $X$; define $Y$ and $Z$ cyclically. Then $\measuredangle B'XC'=\measuredangle C'DB'=\measuredangle CAB$ and $\tfrac{XC'}{XB'}=\tfrac{C'D}{B'D}$ hence $\overline{DX} \perp \overline{B'C'}$. Now observe that $OB=OY$ and $OC=OZ$ so $\overline{OD}$ bisects both angles $BDY$ and $CDZ$ hence $D$ lies on $\overline{YZ}$; and $BYCZ$ is an isosceles trapezoid. Claim: $\overline{YE}, \overline{BH}$ meet at point $H_B$ lying on $\odot(ABC)$. (Proof) Note that $\measuredangle (YE, BH)=\measuredangle (A'C', AC)=\measuredangle CYA'=\measuredangle YCB$ hence proving the claim. $\blacksquare$ Likewise define $H_A, H_C$; then let $T_A=\ell_b \cap \ell_c$. Then $\measuredangle H_BT_AH_C=\measuredangle YT_AZ=\measuredangle (YE, ZF)=\measuredangle (A'C', A'B')=\measuredangle H_BHH_C$ so $H_B, T_A, H, H_C$ are concyclic with center $A$. Finally, $\measuredangle H_BHT_A=\measuredangle H_BH_CZ=\measuredangle (B'C', BH)$ hence proving $\overline{HT_A} \parallel \overline{B'C'} \implies \overline{HT_A} \perp \ell_a$ thus, similar relations show that $H$ is also the orthocenter of $\triangle T_AT_BT_C$ as desired. $\blacksquare$ I did this with geogebra so believing the intermediary claims was a lot easier. The oriented angles might be a bit off; the idea is there though so I'll leave it like this