Giải thích các bước giải:

a.Ta có $\Delta ABC$ vuông tại $A\to BC=\sqrt{AB^2+AC^2}=30$

Mà $CD$ là phân giác $\hat C$

$\to \dfrac{DA}{DB}=\dfrac{CA}{CB}=\dfrac45$

$\to \dfrac{DA}{DA+DB}=\dfrac4{4+5}$

$\to\dfrac{AD}{AB}=\dfrac49$

$\to AD=\dfrac49AB=8\to BD=AB-AD=10$

b.Xét $\Delta BHA, \Delta ABC$ có:

Chung $\hat B$

$\widehat{BHA}=\widehat{BAC}$

$\to \Delta BHA\sim\Delta BAC(g.g)$

c.Xét $\Delta DAC, \Delta DKB$ có

$\widehat{ADC}=\widehat{KDB}$

$\widehat{DAC}=\widehat{DKB}(=90^o)$

$\to \Delta DAC\sim\Delta DKB(g.g)$

$\to\dfrac{DA}{DK}=\dfrac{DC}{DB}$

$\to DA.DB=DK.DC$

d.Gọi $AH\cap CD=G$

Xét $\Delta DAK, \Delta DCB$ có:

$\widehat{KDA}=\widehat{BDC}$

$DK.DC=DB.DA\to \dfrac{DK}{DB}=\dfrac{DA}{DC}$

$\to \Delta DAK\sim\Delta DCB(c.g.c)$

$\to \widehat{KAD}=\widehat{DCB}$

$\to \widehat{KAB}=\widehat{GCH}=90^o-\widehat{HGC}=90^o-\widehat{EGK}=\widehat{GEK}=\widehat{BEA}$

Mà $\widehat{KBA}=\widehat{EBA}$

$\to \Delta BAK\sim\Delta BEA(g.g)$

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

$\to BA^2=BK.BE$

Mà $BA=BF\to BF^2=BK.BE$

$\to \dfrac{BK}{BF}=\dfrac{BF}{BE}$

Mà $\widehat{KBF}=\widehat{EBF}$

$\to \Delta BKF\sim\Delta BFE(c.g.c)$

$\to \widehat{BFE}=\widehat{BKF}=90^o\to BF\perp FE$