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 $\widehat{KBF}=\widehat{EBF}$
$\to \Delta BKF\sim\Delta BFE(c.g.c)$
$\to \widehat{BFE}=\widehat{BKF}=90^o\to BF\perp FE$
