Giải thích các bước giải:

a) $AB^2+AC^2=BC^2$ (ĐL Pytago) 

$\to AC=\sqrt{BC^2-AB^2}=\sqrt{15^2-9^2}=12$

$\dfrac{1}{{A{H^2}}} = \dfrac{1}{{A{B^2}}} + \dfrac{1}{{A{C^2}}} = \dfrac{1}{{{9^2}}} + \dfrac{1}{{{{12}^2}}} \Rightarrow AH = \dfrac{{36}}{5}$

b) Ta có:

$\widehat {AEH} = \widehat {{\rm{AF}}H} = \widehat {EAF} = {90^0}$$\to AEHF$ là hình chữ nhật.

$ \Rightarrow \widehat {AEF} = \widehat {EAH} = \widehat {ACB}$ (Do $\widehat {EAH} ;\widehat {ACB}$ cùng phụ với $\widehat {ABC}$)

Có: $\left\{ \begin{array}{l}
\widehat {EAF} = \widehat {CAB} = {90^0}\\
\widehat {AEF} = \widehat {ACB}
\end{array} \right. \Rightarrow \Delta AEF \sim \Delta ACB(g.g)$

c)

$\left\{ \begin{array}{l}
\widehat {MEB} = \widehat {MCF}\left( { = \widehat {AEF}} \right)\\
\widehat M{\rm{ }}chung
\end{array} \right. \Rightarrow \Delta MEB \sim \Delta MCF(g.g)$

$\begin{array}{l}
 \Rightarrow \dfrac{{ME}}{{MC}} = \dfrac{{MB}}{{MF}}\\
 \Rightarrow ME.MF = MB.MC\\
 \Rightarrow ME.MF = \left( {MI – BI} \right)\left( {MI + IC} \right) = M{I^2} – I{C^2}
\end{array}$

d)

 $\begin{array}{l}
 + )\left\{ \begin{array}{l}
\widehat {BEH} = \widehat {BAC} = {90^0}\\
\widehat B{\rm{ chung}}
\end{array} \right. \Rightarrow \Delta BEH \sim \Delta BAC(g.g)\\
 \Rightarrow \dfrac{{BH}}{{BC}} = \dfrac{{BE}}{{BA}}\\
 \Rightarrow BC.BE = BH.BA\\
 \Rightarrow BC.BE.CF = BH.BA.CF(1)\\
 + )\left\{ \begin{array}{l}
\widehat {AHB} = \widehat {CFH} = {90^0}\\ 
\widehat {ABH} = \widehat {CHF}\left( {HF//AE} \right)
\end{array} \right. \Rightarrow \Delta AHB \sim \Delta CFH(g.g)\\
 \Rightarrow \dfrac{{AH}}{{CF}} = \dfrac{{AB}}{{CH}} \Rightarrow CF.AB = AH.CH(2)\\
(1),(2) \Rightarrow BC.BE.CF = BH.AH.CH = AH.\left( {BH.CH} \right) = A{H^3}
\end{array}$

e)

$ \begin{array}{l}
3A{H^2} + B{E^2} + C{F^2}\\
 = 2A{H^2} + A{H^2} + B{E^2} + C{F^2}\\
 = 2A{H^2}{\rm{ + E}}{{\rm{F}}^2} + B{E^2} + C{F^2}\\
 = 2A{H^2} + H{F^2} + H{E^2} + B{E^2} + C{F^2}\\
 = \left( {A{H^2} + H{E^2} + B{E^2}} \right) + \left( {A{H^2} + H{F^2} + C{F^2}} \right)\\
 = \left( {A{H^2} + H{B^2}} \right) + \left( {A{H^2} + C{H^2}} \right)\\
 = A{B^2} + AC{^2} = B{C^2}
\end{array}$

f)

 $\begin{array}{l}
B{H^3} + C{H^3} + 3.AB.AC.{\rm{EF}}\\
{\rm{ = }}{\left( {BH + CH} \right)^3} – 3BH.CH.\left( {BH + CH} \right) + 3.AH.BC.AH\\
 = B{C^3} – 3A{H^2}.BC + 3.A{H^2}.BC\\
 = B{C^3}
\end{array}$