Giải thích các bước giải:

a.Xét $\Delta OAB,\Delta MEB$ có:

Chung $\hat B$

$\hat O=\hat M(=90^o)$

$\to \Delta OAB\sim\Delta MEB(g.g)$

b.Xét $\Delta AMN,\Delta AOB$ có:
Chung $\hat A$

$\hat M=\hat O(=90^o)$

$\to \Delta AMN\sim\Delta AOB(g.g)$

$\to \dfrac{AM}{AO}=\dfrac{AN}{AB}$

$\to AN.AO=AM.AB$

c.Xét $\Delta AMO,\Delta ANB$ có:
Chung $\hat O$

$\dfrac{AM}{AO}=\dfrac{AN}{AB}$

$\to \Delta AMO\sim\Delta ANB(c.g.c)$

$\to \widehat{AOM}=\widehat{ABN}$

Vì $NE\perp AB, AN\perp BE\to BN\perp AE$

$\to \widehat{NFA}=\widehat{NOB}=90^o)$

$\to \Delta NFA\sim\Delta NOB(g.g)$

$\to \dfrac{NF}{NO}=\dfrac{NA}{NB}$

$\to \Delta NOF\sim\Delta NBA(c.g.c)$

$\to \widehat{NOF}=\widehat{NBA}=\widehat{AOM}$

$\to OA$ là phân giác $\widehat{FOM}$