Giải thích các bước giải:
a.Xét $\Delta BAH,\Delta ABC$ có:
Chung $\hat B$
$\widehat{BHA}=\widehat{BAC}(=90^o)$
$\to\Delta BAH\sim\Delta BCA(g.g)$
b.Xét $\Delta BAD,\Delta BKH$ có:
$\widehat{ABD}=\widehat{KBH}$ vì $BD$ là phân giác $\hat B$
$\widehat{BAD}=\widehat{BHK}(=90^o)$
$\to\Delta BAD\sim\Delta BHK(g.g)$
$\to \dfrac{BA}{BH}=\dfrac{BD}{BK}$
$\to BA.BK=BH.BD$
c.Xét $\Delta ADB,\Delta CDE$ có:
$\widehat{ADB}=\widehat{CDE}$
$\widehat{BAD}=\widehat{DEC}(=90^o)$
$\to\Delta ABD\sim\Delta ECD(g.g)$
$\to \dfrac{DA}{DE}=\dfrac{DB}{DC}$
$\to \dfrac{DA}{DB}=\dfrac{DE}{DC}$
Mà $\widehat{ADE}=\widehat{BDC}$
$\to\Delta DAE\sim\Delta DBC(c.g.c)$
$\to \widehat{DAE}=\widehat{DBC}\to \widehat{EAC}=\widehat{DBC}=\widehat{ABD}=\widehat{ECD}=\widehat{ECA}$
$\to\Delta ACE$ cân tại $E\to EA=EC$
