Giải thích các bước giải:
a.Xét $\Delta ABH,\Delta ABE$ có:
Chung $\hat A$
$\widehat{AHB}=\widehat{ABE}(=90^o)$
$\to \Delta AHB\sim\Delta ABE(g.g)$
$\to \dfrac{AB}{AE}=\dfrac{AH}{AB}\to AB^2=AH\cdot AE$
b.Xét $\Delta HBE, \Delta HAB$ có:
$\widehat{AHB}=\widehat{BHE}(=90^o)$
$\widehat{HBE}=90^o-\widehat{HBA}=\widehat{HAB}$
$\to \Delta HBE\sim\Delta HAB(g.g)$
$\to \dfrac{HB}{HA}=\dfrac{HE}{HB}$
$\to HB^2=HA.HE$
c.Ta có: $AD//BC$
$\to \dfrac{HA}{HE}=\dfrac{HD}{HB}=\dfrac{HF}{HA}$ vì $AB//CD$
$\to HA^2=HE.HF$