Đáp án:

$\begin{array}{l}
a)Xet:\Delta AMH;\Delta AHB:\\
 + )\widehat {AMH} = \widehat {AHB} = {90^0}\\
 + )\widehat {MAH}\,chung\\
 \Rightarrow \Delta AMH \sim \Delta AHB\left( {g - g} \right)\\
 \Rightarrow \dfrac{{AM}}{{AH}} = \dfrac{{AH}}{{AB}}\\
 \Rightarrow AM.AB = A{H^2}
\end{array}$

Tương tự ta chứng minh được AN.AC=AH^2

=> AM.AB=AN.AC

$\begin{array}{l}
b)Xét:\Delta AMH;\Delta HMB:\\
 + )\widehat {AMH} = \widehat {HMB} = {90^0}\\
 + )\widehat {MAH} = \widehat {MHB}\left( { + \widehat {MBH} = {{90}^0}} \right)\\
 \Rightarrow \Delta AMH \sim \Delta HMB\left( {g - g} \right)\\
 \Rightarrow \dfrac{{MA}}{{MH}} = \dfrac{{MH}}{{MB}}\\
 \Rightarrow MA.MB = M{H^2}\\
TT:NA.NC = N{H^2}\\
 \Rightarrow MA.MB + NA.NC = M{H^2} + N{H^2}\\
 \Rightarrow MA.MC + NA.NC = M{N^2} = A{H^2}\\
Do:\Delta AHB \sim \Delta CHA\left( {g - g} \right)\\
 \Rightarrow \dfrac{{HA}}{{HC}} = \dfrac{{HB}}{{HA}}\\
 \Rightarrow HB.HC = H{A^2}\\
 \Rightarrow HB.HC = MA.MC + NA.NC\\
c)Do:\Delta AHB \sim \Delta CHA\\
 \Rightarrow \dfrac{{AB}}{{AC}} = \dfrac{{HB}}{{HA}} = \dfrac{{HA}}{{HC}}\\
 \Rightarrow \dfrac{{HB}}{{HA}}.\dfrac{{HA}}{{HC}} = \dfrac{{AB}}{{AC}}.\dfrac{{AB}}{{AC}}\\
 \Rightarrow \dfrac{{HB}}{{HC}} = {\left( {\dfrac{{AB}}{{AC}}} \right)^2}
\end{array}$