Đáp án:

$\begin{array}{l}
a)Theo\,Pytago:\\
A{C^2} = D{A^2} + D{C^2}\\
 = {5^2} + {12^2} = 169\\
 \Rightarrow AC = 13\left( {cm} \right)\\
Theo\,t/c:\\
\dfrac{{AI}}{{DA}} = \dfrac{{CI}}{{DC}}\\
 \Rightarrow \dfrac{{AI}}{{12}} = \dfrac{{CI}}{5} = \dfrac{{AI + CI}}{{12 + 5}} = \dfrac{{13}}{{17}}\\
 \Rightarrow \left\{ \begin{array}{l}
AI = \dfrac{{156}}{{17}}\left( {cm} \right)\\
CI = \dfrac{{65}}{{17}}\left( {cm} \right)
\end{array} \right.\\
b)Xet:\Delta FAH;\Delta DCH:\\
 + \widehat {HFA} = \widehat {HDC} = {90^0}\\
 + \widehat {FHA} = \widehat {DHC}\\
 \Rightarrow \Delta FAH \sim \Delta DCH\left( {g - g} \right)\\
 \Rightarrow \widehat {FAH} = \widehat {DCH}\\
hay\,\widehat {BAD} = \widehat {BCF}\\
 \Rightarrow \Delta BAD \sim \Delta BCF\left( {g - g} \right)\\
 \Rightarrow \dfrac{{BA}}{{BC}} = \dfrac{{BD}}{{BF}}\\
 \Rightarrow BD.BC = BA.BF\\
c)\Delta DCH \sim FCD\left( {g - g} \right)\\
 \Rightarrow \dfrac{{DC}}{{CH}} = \dfrac{{CF}}{{DC}}\\
 \Rightarrow D{C^2} = CH.CF\\
TT:D{A^2} = AH.AD\\
 \Rightarrow CH.CF + AH.AH = D{C^2} + D{A^2} = A{C^2}
\end{array}$