Let $D$ $\equiv$ $BC_B$ $\cap$ $CB_C$, $E$ $\equiv$ $AC_A$ $\cap$ $CA_C$, $F$ $\equiv$ $AB_A$ $\cap$ $BA_B$; $d_A$, $d_B$, $d_C$ be the lines through $A$, $B$, $C$ and perpendicular with $B_AC_A$, $C_BA_B$, $A_CB_C$, respectively We have: $\angle{ABA_B} = \angle{BAB_A} = \angle{ACA_C} = \angle{CAC_A} = 180^o - 2 \alpha$ Then: $\triangle ABF$ $\sim$ $\triangle ACE$ So: $\dfrac{AF}{AB_A} = \dfrac{AF}{AB} = \dfrac{AE}{AC} = \dfrac{AE}{AC_A}$ or $EF$ $\parallel$ $B_AC_A$ Similarly: $FD$ $\parallel$ $C_BA_B$, $DE$ $\parallel$ $A_CB_C$ But: $AE^2 - AF^2 + BF^2 - BD^2 + CD^2 - CE^2 = 0$ then by Carnot theorem, we have: $d_A$, $d_B$, $d_C$ are concurrent or the lines through $A$, $B$, $C$ and perpendicular with $B_AC_A$, $C_BA_B$, $A_CB_C$ concurrent

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