Giải thích các bước giải:
a.Ta có:
$\widehat{AHB}=\widehat{AHC}=90^o,\widehat{BAH}=90^o-\widehat{HAC}=\widehat{ACH}$
$\to \Delta AHB\sim\Delta CHA(g.g)$
$\to\dfrac{AH}{CH}=\dfrac{HB}{HA}$
$\to AH^2=HB.HC=36\to AH=6$
$\to AB=\sqrt{AH^2+BH^2}=2\sqrt{13}, AC=\sqrt{AH^2+CH^2}=3\sqrt{13}$
b.Vì $BD$ là phân giác góc $B\to \widehat{ABD}=\widehat{EBH}$
Mà $\widehat{BAD}=\widehat{BHE}=90^o$
$\to \Delta BHE\sim\Delta BAD(g.g)$
$\to \dfrac{S_{EBH}}{S_{BDA}}=(\dfrac{BH}{BA})^2=\dfrac{4}{13}$
c.Ta có $BD$ là phân giác góc $B$
$\to\dfrac{EA}{EH}=\dfrac{BA}{BH}, \dfrac{DA}{DC}=\dfrac{BA}{BC}$
Mà $\widehat{BHA}=\widehat{BAC}=90^o, \widehat{ABH}=\widehat{ABC}$
$\to \Delta BAH\sim\Delta BCA(g.g)$
$\to \dfrac{BA}{BC}=\dfrac{BH}{BA}$
$\to \dfrac{DA}{DC}=\dfrac{EH}{EA}$
$\to EA.DA=EH.DC$
