a)

Xét $\Delta BHA$ và $\Delta AHC$, ta có:

$\widehat{BHA}=\widehat{AHC}=90{}^\circ $

$\widehat{ABH}=\widehat{CAH}$ ( cùng phụ $\widehat{ACB}$ )

$\to \Delta BHA\backsim\Delta AHC\,\,\,\left( g.g \right)$

$\to \dfrac{BH}{AH}=\dfrac{AH}{HC}$

$\to A{{H}^{2}}=BH.HC$

c)

$\Delta ABD\backsim\Delta HBE\,\,\,\left( cmt \right)$

$\to \dfrac{EH}{AD}=\dfrac{EB}{DB}\,\,\,\,\,\left( 1 \right)$

Xét $\Delta ABE$ và $\Delta CBD$, ta có:

$\widehat{ABE}=\widehat{CBD}$ ( vì $BD$ là tia phân giác $\widehat{ABC}$ )

$\widehat{BAE}=\widehat{BCD}$ ( cùng phụ $\widehat{ABC}$ )

$\to \Delta ABE\backsim\Delta CBD\,\,\,\left( g.g \right)$

$\to \dfrac{EA}{DC}=\dfrac{EB}{DB}\,\,\,\,\,\left( 2 \right)$

Từ $\left( 1 \right)$ và $\left( 2 \right)$, ta được:

$\,\,\,\,\,\,\,\dfrac{EH}{AD}=\dfrac{EA}{DC}$

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