a)
$AC=\sqrt{B{{C}^{2}}-A{{B}^{2}}}=\sqrt{{{25}^{2}}-{{15}^{2}}}=20cm$
$\sin B=\dfrac{AC}{BC}=\dfrac{20}{25}=\dfrac{4}{5}\Rightarrow \widehat{B}=53{}^\circ 8'$
b)
Có $\begin{cases}AH^2=MA.AB\\AH^2=AN.AC\end{cases}\Rightarrow MA.AB=AN.AC$
c)
Có $\begin{cases}AC^2=CH.BC\\AB^2=BH.BC\end{cases}$
$\Leftrightarrow\dfrac{AC^2}{AB^2}=\dfrac{CH}{BH}$
$\Leftrightarrow\dfrac{AC^4}{AB^4}=\dfrac{CH^2}{BH^2}$
$\Leftrightarrow\dfrac{AC^4}{AB^4}=\dfrac{CN.AC}{BM.AB}$
$\Leftrightarrow\dfrac{AC^3}{AB^3}=\dfrac{CN}{BM}$
