Cercul

Coordonatele baricentrice absolute ale punctului $A'$ în raport cu triunghiul $ABC$ satisfac atât ecuația medianei din $A$, cât și ecuația cercului:

$$\begin{rcases} x_1+y_1+z_1=1 \\ y_1=z_1 \\ a^2y_1z_1 + b^2x_1z_1 + c^2x_1y_1 = 0 \end{rcases} \Rightarrow \boxed{A'\bigg(\dfrac{-a^2}{2(b^2+c^2)-a^2},~ \dfrac{b^2+c^2}{2(b^2+c^2)-a^2},~ \dfrac{b^2+c^2}{2(b^2+c^2)-a^2}\bigg)}.$$

Analog, $\boxed{B'\bigg(\dfrac{c^2+a^2}{2(c^2+a^2)-b^2},~ \dfrac{-b^2}{2(c^2+a^2)-b^2},~ \dfrac{c^2+a^2}{2(c^2+a^2)-b^2}\bigg)}$ și

$\boxed{C'\bigg(\dfrac{a^2+b^2}{2(a^2+b^2)-c^2},~ \dfrac{a^2+b^2}{2(a^2+b^2)-c^2},~ \dfrac{-c^2}{2(a^2+b^2)-c^2}\bigg)}.$

"$\Rightarrow$": $a=b=c \Rightarrow A'\bigg(-\dfrac{1}{3},~\dfrac{2}{3},~\dfrac{2}{3}\bigg), B'\bigg(\dfrac{2}{3},~-\dfrac{1}{3},~\dfrac{2}{3}\bigg),~ C'\bigg(\dfrac{2}{3},~\dfrac{2}{3},~-\dfrac{1}{3}\bigg) \Rightarrow G'\bigg(\dfrac{1}{3},~\dfrac{1}{3},~\dfrac{1}{3}\bigg)$, deci $\boxed{G'=G}.$

"$\Leftarrow$": Dacă $G'=G$, atunci $\dfrac{x_1+x_2+x_3}{3}=\dfrac{1}{3}$ și analoagele. Folosind formula medianei, obținem:

$\dfrac{-a^2}{m_a^2} + \dfrac{c^2+a^2}{m_b^2} + \dfrac{a^2+b^2}{m_c^2} = 4 \quad (1)$

$\dfrac{b^2+c^2}{m_a^2} + \dfrac{-b^2}{m_b^2} + \dfrac{a^2+b^2}{m_c^2} = 4 \quad (2)$

$\dfrac{b^2+c^2}{m_a^2} + \dfrac{c^2+a^2}{m_b^2} + \dfrac{-c^2}{m_c^2} = 4 \quad (3)$

$$\begin{rcases} (1)-(2) \Rightarrow m_a=m_b \\ (2)-(3) \Rightarrow m_b=m_c \end{rcases} \Rightarrow m_a=m_b=m_c \Rightarrow \boxed{a=b=c}.$$