My solution: Let $D,E,F$ be midpoint of $BC,CA,AB$. $A_1,B_1,C_1$ are circumcenters of $\triangle OBC,\triangle OCA,\triangle OAB$. Since $\overline{AH}.\overline{AA'}=AO^2$, $\triangle AHO$ and $\triangle AOA'$ are similar $\Rightarrow OA'/OH=OA/HA$ Similar ly, $OB'/OH=OB/HB$ $\Rightarrow OA'/OB'=HB/HA=\cos B /\cos C=OE/OD$ $\angle A'OB'=\angle A'OH +\angle B'OH = \angle AHO -\angle AOH +\angle BHO -\angle BOH$ $=360^o -\angle AHB - \angle AOB =180^o +\angle ACB -2\angle ACB = \angle AHB=\angle DOE$ $\Rightarrow \triangle OED \sim \triangle OA'B' \sim \triangle OA_1B_1$ Similarly $\triangle OB_1C_1 \sim \triangle OB'C' ,\triangle OC_1A_1\sim\triangle OC'A'$ $\Longrightarrow A_1B_1C_1O\sim A'B'C'O$ Since $O$ is incenter of $\triangle A_1B_1C_1$, then $O$ is incenter of $\triangle A'B'C'$ Remark. $\angle A'OB'=\angle AHB=\angle A'HB'$... $\Rightarrow$ $H,O$ are antigonal conjugate WRT $\triangle A'B'C'$ $\Rightarrow$ Nine-point center $N$ of $\triangle ABC$ is Feuerbach point of $\triangle A'B'C'$

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I have an idea Let $ABC$ be a triangle and $P,Q$ are two isogonal conjugate points. Circle passes through $P,Q$ and is tangent to $QA$ which cuts $PA$ at $D$. Similarly, we have $E,F$. Let $XYZ$ be pedal triangle of $P$. a) Prove that $\triangle XYZ\cup P\sim\triangle DEF\cup Q$. b) Let $R$ be isogonal conjugate of $Q$ with respect to triangle $DEF$. Prove that $R$ and $P$ are anitgonal conjugate with respect to triangle $DEF$.

buratinogigle wrote: I have an idea Let $ABC$ be a triangle and $P,Q$ are two isogonal conjugate points. Circle passes through $P,Q$ and is tangent to $QA$ which cuts $PA$ at $D$. Similarly, we have $E,F$. Let $XYZ$ be pedal triangle of $P$. a) Prove that $\triangle XYZ\cup P\sim\triangle DEF\cup Q$. From symmetry, it suffices to prove $ \triangle PYZ $ $ \stackrel{-}{\sim} $ $ \triangle QEF $. From $ \triangle BPQ $ $ \sim $ $ \triangle BQE $ and $ \triangle CPQ $ $ \sim $ $ \triangle CQF $ we get $ EQ $ $ = $ $ PQ $ $ \cdot $ $ (BQ/BP) $ and $ FQ $ $ = $ $ PQ $ $ \cdot $ $ (CQ/CP) $, so $ EQ/FQ $ $ = $ $ (CP/BP) $ $ \cdot $ $ (BQ/CQ) $. On the other hand, let $ \triangle Q_aQ_bQ_c $ be the pedal triangle of $ Q $ WRT $ \triangle ABC $, then notice $ \triangle BQQ_c $ $ \sim $ $ \triangle BPX $ and $ \triangle CQQ_b $ $ \sim $ $ \triangle CPX $ we get $ (YP/ZP) $ $ = $ $ (QQ_c/QQ_b) $ $ = $ $ (CP/BP) $ $ \cdot $ $ (BQ/CQ) $ $ = $ $ EQ/FQ $, so combine $ \measuredangle FQE $ $ = $ $ \measuredangle FQC $ $ + $ $ \measuredangle CQB $ $ + $ $ \measuredangle BQE $ $ = $ $ \measuredangle CPQ $ $ + $ $ \measuredangle CQB $ $ + $ $ \measuredangle QPB $ $ = $ $ \measuredangle CPB $ $ + $ $ \measuredangle CQB $ $ = $ $ \measuredangle CAB $ $ = $ $ \measuredangle YPZ $ we conclude that $ \triangle PYZ $ $ \stackrel{-}{\sim} $ $ \triangle QEF $. buratinogigle wrote: b) Let $R$ be isogonal conjugate of $Q$ with respect to triangle $DEF$. Prove that $R$ and $P$ are anitgonal conjugate with respect to triangle $DEF$. From symmetry, it suffices to prove the reflection $ R_d $ of $ R $ in $ EF $ lie on $ \odot (EPF) $. Since $ \measuredangle ER_dF $ $ = $ $ \measuredangle FRE $ $ = $ $ \measuredangle DFQ $ $ + $ $ \measuredangle QED $ $ = $ $ \measuredangle PZX $ $ + $ $ \measuredangle XYP $ $ = $ $ \measuredangle PBC $ $ + $ $ \measuredangle BCP $ $ = $ $ \measuredangle BPC $ $ = $ $ \measuredangle EPF $, so we get $ R_d $ $ \in $ $ \odot (EPF) $.

Remark : Here is a nice equivalent property of the generalization mentioned by buratinogigle at post #5 : Given $ \triangle ABC $ and a point $ P $. Let $ Q $ be the image of $ P $ under the Inversion $ \mathbf{I}(\odot (ABC)) $. Let $ \triangle P_aP_bP_c, $ $ \triangle Q_aQ_bQ_c $ be the pedal triangle of $ P, $ $ Q $ WRT $ \triangle ABC $, respectively. Let $ M $ be the midpoint of $ PQ $ and $ R $ be the isogonal conjugate of $ P $ WRT $ \triangle P_aP_bP_c $. Then $ \triangle P_aP_bP_c $ $ \cup $ $ R $ $ \stackrel{-}{\sim} $ $ \triangle Q_aQ_bQ_c $ $ \cup $ $ M $.

P.S

Complex number also works well for this problem，which perfectly handles various configurations. PS. $\triangle ABC$ should be acute, otherwise $O$ must be one of excenters of $\triangle A'B'C'$.

jred wrote: Complex number also works well for this problem，which perfectly handles various configurations. PS. $\triangle ABC$ should be acute, otherwise $O$ must be one of excenters of $\triangle A'B'C'$. we solve the problem using complex number and we take $\odot ABC$ the unit circle! because $AO^2=AH.AA' \implies \triangle AHO \stackrel{-}{\sim} AA'O \implies \frac{a-0}{a-h}=\frac{\overline{a}-\overline{a'}}{\overline{a}-0} \implies a'=\frac{\sum ab}{b+c}$ and similarly $ b'=\frac{\sum ab}{a+c}, c'=\frac{\sum ab}{b+a}$ and now we prove $O$ has the same distance from sides of $\triangle A'B'C'$ in order to prove this we prove the $d(O,AC)$ is symmetric with respect to $a,b,c$ we know that: $d(O,A'C').A'C'=S(OA'C')$ and with the areal formula it implies: $\begin{vmatrix} 0 & 0 &1 \\ a' & \overline{a'} &1 \\ c'& \overline{c'} & 1 \\ \end{vmatrix}=|a'-c'|.d(O,A'C')$ and the rest is just 5 minutes of computing!