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 $\widehat{BHA}=\widehat{BAC}=90^o, \widehat{ABH}=\widehat{ABC}$

$\to \Delta BAH\sim\Delta BCA(g.g)$

$\to \dfrac{BA}{BC}=\dfrac{BH}{BA}$

$\to \dfrac{DA}{DC}=\dfrac{EH}{EA}$

$\to EA.DA=EH.DC$