Giải thích các bước giải:
a.Xét $\Delta ABE,\Delta HBE$ có:
$\widehat{ABE}=\widehat{HBE}$
Chung $BE$
$\widehat{EAB}=\widehat{EHB}(=90^o)$
$\to \Delta ABE=\Delta HBE$(cạnh huyền-góc nhọn)
b.Từ câu a $\to EA=EH$
Xét $\Delta AEK,\Delta HEC$ có:
$\widehat{EAK}=\widehat{EHC}(=90^o)$
$EA=EH$
$\widehat{AEK}=\widehat{HEC}$
$\to\Delta AEK=\Delta HEC(g.c.g)$
$\to EK=EC$
c.Từ câu a $\to BA=BH$
Xét $\Delta BHK,\Delta BAC$ có:
Chung $\hat B$
$BH=BA$
$\widehat{BHK}=\widehat{BAC}(=90^o)$
$\to\Delta BHK=\Delta BAC(g.c.g)$
$\to HK=AC<CB$