Giải thích các bước giải:
a.Ta có $\Delta ABC$ vuông tại $A$
$\to BC^2=AB^2+AC^2=100\to BC=10$
b.Xét $\Delta AHB, \Delta AHD$ có:
Chung $AH$
$\widehat{AHB}=\widehat{AHD}(=90^o)$
$HB=HD$
$\to \Delta AHB=\Delta AHD(c.g.c)$
$\to AB=AD$
c.Xét $\Delta AHB,\Delta EHD$ có:
$HA=HE$
$\widehat{AHB}=\widehat{DHE}$
$HB=HD$
$\to\Delta AHB=\Delta EHD(c.g.c)$
$\to \widehat{ABH}=\widehat{DHE}$
$\to AB//DE$
Mà $AB\perp AC\to DE\perp AC$
d.Ta có $ AB<AC\to \widehat{ABC}>\widehat{ACB}$
Mà $\widehat{ABC}=\widehat{ABH},\widehat{BAH}=90^o-\widehat{ABH}=90^o-\widehat{ABC}=\widehat{ACB}$
$\to \widehat{ABH}>\widehat{BAH}$
$\to BH<AH$
$\to 2BH<2AH$
$\to BD<AE$