Giải thích các bước giải:
a.Xét $\Delta AHF,\Delta ABD$ có:
Chung $\hat A$
$\widehat{AFH}=\widehat{ADB}(=90^o)$
$\to \Delta AFH\sim\Delta ADB(g.g)$
$\to\dfrac{AF}{AD}=\dfrac{AH}{AB}$
$\to AF.AB=AH.AD$
Tương tự $AE.AC=AH.AD$
$\to AE.AC=AH.AD=AF.AB$
b.Xét $\Delta HAE,\Delta HBD$ có:
$\widehat{AHE}=\widehat{BHD}$
$\widehat{AEH}=\widehat{HDB}(=90^o)$
$\to \Delta AHE\sim\Delta BHD(g.g)$
$\to \dfrac{AH}{BH}=\dfrac{HE}{HD}$
$\to HA.HD=HE.HB$
Tương tự $HA.HD=HF.HC$
$\to HA.HD=HE.HB=HC.HF$
c.Xét $\Delta AFD,\Delta ABH$ có:
Chung $\hat A$
$\dfrac{AF}{AH}=\dfrac{AD}{AB}$ vì $AF.AB=AH.AD$
$\to\Delta AFD\sim\Delta AHB(c.g.c)$
$\to \widehat{ADF}=\widehat{ABH}=\widehat{FBH}$
Tương tự $\widehat{ADE}=\widehat{HCE}$
Mà $\widehat{HBF}=90^o-\widehat{FHB}=90^o-\widehat{EHC}=\widehat{ECH}$
$\to \widehat{ADF}=\widehat{ADE}$
$\to DA$ là phân giác $\widehat{EDF}$
