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خط دلخواه 
$\ell$ 
از 
$O$ 
مرکز ارتفاعی مثلث دلخواه 
$ABC$ 
عبور می کند. نقطه دلخواه 
$P$ 
روی دایره محیطی 
$\triangle ABC$ 
قرار دارد. خطوط 
$AP$ ، $BP$ 
و 
$CP$ 
به ترتیب خط 
$\ell$ 
را در 
$A_P$ ، $B_P$ 
و 
$C_P$ 
قطع می کنند. فرض کنید 
$A_0$ ، $B_0$ 
و 
$C_0$ 
به ترتیب تصاویر 
$A_P$ ، $B_P$ 
و 
$C_P$ 
روی اضلاع 
$BC$ ، $CA$ 
و 
$AB$ 
باشند. ثابت کنید 
$A_0$ ، $B_0$ 
و 
$C_0$ 
هم خط اند و خط گذرنده از این سه نقطه پاره خط واصل 
$P$ 
و مرکز ارتفاعی مثلث 
$ABC$ 
را نصف می کند.

\vspace{.5cm}

\textbf{
راه حل:
}

%

برای هر نقطه 
$X$ 
روی دایره محیطی 
$ABC$ 
فرض کنید 
$\ell_X$ 
معرف خط سیمسون 
$X$ 
باشد.

%

فرض کنید 
$H$ 
مرکز ارتفاعی مثلث 
$ABC$ ، $Q$ 
قرینه 
$P$ 
نسبت به 
$\ell$ ، $M$ 
وسط 
$PH$ 
و 
$A'$ 
قطری 
$A$ 
باشد. 
$P_A$ 
و 
$H_A$ 
به ترتیب تصاویر 
$P$ 
و 
$A$ 
روی 
$BC$ 
هستند. توجه داشته باشید:

\vspace{.5cm}

\begin{LTR}
$\left.\begin{array}{lll}
OP=OA\ ,\ MH_A=MP_A\\
\angle(MP_A,P_AH_A)=\angle(\ell_P,\ell_{A'})=\dfrac{PA'}{2}=\angle(OA,AP)\\
\dfrac{A_PA}{A_PP}=\dfrac{A_0H_A}{A_0P_A}
\end{array}\right\}\Longrightarrow \triangle MP_AH_A\cup A_0\sim \triangle OPA\cup A_P$
\end{LTR}

\vspace{.5cm}

بنابرین:
$$\angle(\ell_P,MA_0)=\angle(P_AM,MA_0)=\angle(PO,OA_P)=\dfrac{\widearc{PQ}}{2}=\angle(\ell_P,\ell_Q)\Longrightarrow \ell_Q\parallel MA_0$$
به طریق مشابه ثابت می شود:
$$MB_0\parallel \ell_Q\ ,\ MC_0\parallel \ell_Q$$
در نتیجه نقاط 
$A_0$ ، $B_0$ ، $C_0$ 
و 
$M$ 
هم خط اند.
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\end{document}