Gọi $I = AM\cap EF$
Ta có:
$MA//NB//OC \, (\perp EF)$
Áp dụng định lý $Thales$ ta được:
$+) \quad \dfrac{IC}{IB} = \dfrac{IO}{IN}$
$\Rightarrow \dfrac{IC}{BC} = \dfrac{IO}{ON}$ $(1)$
$+) \quad \dfrac{IC}{IA} = \dfrac{IO}{IM}$
$\Rightarrow \dfrac{IC}{AC} = \dfrac{IO}{OM}$ $(2)$
$(1)(2) \Rightarrow \dfrac{BC}{AC} = \dfrac{ON}{OM}$
mà $OM = ON = R$
nên $\dfrac{BC}{AC} = 1$
$\Rightarrow BC = AC$
Ta lại có:
$OC\perp EF$
$\Rightarrow EC = CF = \dfrac{1}{2}EF$
$\Rightarrow EC - BC = CF - AC$
$\Rightarrow BE = AF$
$\Rightarrow BE + AB = AF + AB$
$\Rightarrow AE = BF$
