Ta có: \(\angle ACO+\angle ABO=90+90=180\Rightarrow ABOC\) nội tiếp
Xét \(\Delta HAC\) và \(\Delta HBO:\) Ta có: \(\left\{{}\begin{matrix}\angle HAC=\angle HBO\\\angle AHC=\angle BHO\end{matrix}\right.\)
\(\Rightarrow\Delta HAC\sim\Delta HBO\left(g-g\right)\Rightarrow\dfrac{HA}{HB}=\dfrac{HC}{HO}\Rightarrow HA.HO=HB.HC\)
Xét \(\Delta EHC\) và \(\Delta BHF:\) Ta có: \(\left\{{}\begin{matrix}\angle EHC=\angle BHF\\\angle CEH=\angle FBH\end{matrix}\right.\)
\(\Rightarrow\Delta EHC\sim\Delta BHF\left(g-g\right)\Rightarrow\dfrac{HE}{HB}=\dfrac{HC}{HF}\Rightarrow HB.HC=HE.HF\)
\(\Rightarrow HE.HF=HA.HO\Rightarrow\dfrac{HE}{HA}=\dfrac{HO}{HF}\)
Xét \(\Delta EHO\) và \(\Delta AHF:\) Ta có: \(\left\{{}\begin{matrix}\angle EHO=\angle AHF\\\dfrac{HE}{HA}=\dfrac{HO}{HF}\end{matrix}\right.\)
\(\Rightarrow\Delta EHO\sim\Delta AHF\left(c-g-c\right)\Rightarrow\angle OEH=\angle HAF\)
\(\Rightarrow AEOF\) nội tiếp