$\widehat{C_1}=\widehat {C_2}$ (đối đỉnh)
$\widehat{C_2}=\widehat{B_2}$ ($\Delta ABC$ cân đỉnh $A$)
$\Rightarrow \widehat{C_1}=\widehat{B_2}$
$\widehat{E_1}=\widehat{B_1}$ ($\Delta OEB$ cân đỉnh $O$)
Mà $\widehat{E_1}+\widehat{C_1}=\widehat{B_1}+\widehat{B_2}=90^o$
$\Rightarrow \widehat{EOC}=90^o$
$\Rightarrow AO\bot EO$