Giải thích các bước giải:

 a) Ta có:

$\begin{array}{l}
\Delta ABC;\widehat {BAC} = {90^0};AB = 15cm;AC = 20cm\\
 \Rightarrow \left\{ \begin{array}{l}
BC = \sqrt {A{B^2} + A{C^2}}  = \sqrt {{{15}^2} + {{20}^2}}  = 25cm\\
\dfrac{1}{{A{H^2}}} = \dfrac{1}{{A{B^2}}} + \dfrac{1}{{A{C^2}}} \Rightarrow AH = 12cm\\
A{B^2} = BH.BC \Rightarrow BH = \dfrac{{A{B^2}}}{{BC}} = 9cm
\end{array} \right.
\end{array}$

b) Ta có:

$AM$ là phân giác $\widehat {BAC}$

Khi đó ta có:

$\begin{array}{l}
\dfrac{{BM}}{{CM}} = \dfrac{{AB}}{{AC}} = \dfrac{{15}}{{20}} = \dfrac{3}{4};BM + CM = BC = 25\left( {cm} \right)\\
 \Rightarrow BM = \dfrac{{75}}{7}cm;CM = \dfrac{{100}}{7}cm\\
 \Rightarrow HM = BM - BH = \dfrac{{75}}{7} - 9 = \dfrac{{12}}{7}cm
\end{array}$

Lại có:

$\begin{array}{l}
\Delta AHM;\widehat {AHM} = {90^0};AH = 12cm;HM = \dfrac{{12}}{7}cm\\
 \Rightarrow AM = \sqrt {A{H^2} + H{M^2}}  = \dfrac{{60\sqrt 2 }}{7}cm
\end{array}$

Vậy $AM = \dfrac{{60\sqrt 2 }}{7}cm$

d) Ta có:

$\begin{array}{l}
\left\{ \begin{array}{l}
HD//AC\left( { \bot AB} \right) \Rightarrow \dfrac{{BD}}{{BA}} = \dfrac{{HD}}{{CA}} \Rightarrow BD = \dfrac{{AB}}{{AC}}.HD\\
HE//AB\left( { \bot AC} \right) \Rightarrow \dfrac{{CE}}{{CA}} = \dfrac{{HE}}{{BA}} \Rightarrow CE = \dfrac{{AC}}{{AB}}.HE
\end{array} \right.\\
 \Rightarrow \dfrac{{BD}}{{CE}} = {\left( {\dfrac{{AB}}{{AC}}} \right)^2}.\dfrac{{HD}}{{HE}}\left( 1 \right)
\end{array}$

Lại có:

$\widehat {DAE} = \widehat {ADH} = \widehat {AEH} = {90^0}$

$\to ADHE$ là hình chữ nhật.

$\begin{array}{l}
 \Rightarrow \widehat {DHE} = {90^0};\widehat {HDE} = \widehat {HAC}\\
 \Rightarrow \widehat {DHE} = {90^0};\widehat {HDE} = \widehat {ABC}
\end{array}$

Khi đó:

$\begin{array}{l}
\left\{ \begin{array}{l}
\widehat {DHE} = \widehat {BAC} = {90^0}\\
\widehat {HDE} = \widehat {ABC}
\end{array} \right. \Rightarrow \Delta DHE \sim \Delta BAC\left( {g.g} \right)\\
 \Rightarrow \dfrac{{HD}}{{AB}} = \dfrac{{HE}}{{AC}} \Rightarrow \dfrac{{HD}}{{HE}} = \dfrac{{AB}}{{AC}}\left( 2 \right)
\end{array}$

Từ (1),(2) $ \Rightarrow \dfrac{{BD}}{{CE}} = {\left( {\dfrac{{AB}}{{AC}}} \right)^3}$