Giải thích các bước giải:

a.Xét $\Delta CAK,\Delta CFB$ có:
Chung $\hat C$

$\hat K=\hat F(=90^o)$

$\to \Delta CKA\sim\Delta CFB(g.g)$

Xét $\Delta BEC,\Delta BAK$ có:

Chung $\hat B$

$\hat K=\hat E(=90^o)$

$\to \Delta BEC\sim\Delta BKA(g.g)$

$\to \dfrac{BE}{BK}=\dfrac{BC}{BA}$

$\to BE.BA=BK.BC$

b.Từ a $\to \dfrac{CK}{CF}=\dfrac{CA}{CB}$

Mà $\widehat{KCF}=\widehat{ACB}$

$\to \Delta CKF\sim\Delta CAB(c.g.c)$

c.Từ b $\to  \dfrac{CK}{CF}=\dfrac{CA}{CB}$

$\to CF.CA=CK.CB$

$\to BE.BA+CF.CA=BK.BC+CK.BC=BC^2=81$