Giải thích các bước giải:
a.Xét $\Delta ABE,\Delta ACF$ có:
Chung $\hat A$
$\widehat{AEB}=\widehat{AFC}(=90^o)$
$\to\Delta ABE\sim\Delta ACF(g.g)$
b.Ta có $BE\cap CF=O\to O$ là trực tâm $\Delta ABC\to AM\perp BC$
Xét $\Delta COM,\Delta CFB$ có:
Chung $\hat C$
$\widehat{CMO}=\widehat{CFB}=90^o$
$\to\Delta CMO\sim\Delta CFB(g.g)$
$\to \dfrac{CM}{CF}=\dfrac{CO}{CB}$
$\to \dfrac{CM}{CO}=\dfrac{CF}{CB}$
Mà $\widehat{FCM}=\widehat{OCB}$
$\to \Delta CFM\sim\Delta CBO(c.g.c)$
$\to \widehat{CFM}=\widehat{CBO}$
Tương tự $\widehat{CFE}=\widehat{CAO}$
Mà $\widehat{CBO}=\widehat{MBO}=90^o-\widehat{BOM}=90^o-\widehat{AOE}=\widehat{OAE}=\widehat{CAO}$
$\to \widehat{EFC}=\widehat{MFC}$
$\to FC$ là phân giác $\widehat{EFM}$
c.Từ câu a $\to \dfrac{AE}{AB}=\dfrac{AF}{AC}=\dfrac{BE}{CF}$
Mà $S_{ABC}=\dfrac12BE.AC=\dfrac12CF.AB$
$\to \dfrac{BE}{CF}=\dfrac{AB}{AC}$
$\to \dfrac{BE}{AB}=\dfrac{CF}{AC}$
Để $AB.AC=AE.AF+BE.CF$
$\to\dfrac{AE.AF}{AB.AC}+\dfrac{BE.CF}{AB.AC}=1$
$\to\dfrac{AE}{AB}.\dfrac{AF}{AC}+\dfrac{BE}{AB}.\dfrac{CF}{AC}=1$
$\to (\dfrac{AE}{AB})^2+(\dfrac{BE}{AB})^2=1$
$\to \dfrac{AE^2+BE^2}{AB^2}=1$ đúng vì $\Delta ABE$ vuông tại $E$
