Đáp án:
$\begin{array}{l}
12\widehat A = 10\widehat B = 15\widehat C\\
\Rightarrow \frac{{12\widehat A}}{{60}} = \frac{{10\widehat B}}{{60}} = \frac{{15\widehat C}}{{60}} = \frac{{\widehat A}}{5} = \frac{{\widehat B}}{6} = \frac{{\widehat C}}{4}\\
\Rightarrow \frac{{\widehat A + \widehat B + \widehat C}}{{5 + 6 + 4}} = \frac{{{{180}^0}}}{{15}} = {12^0}\\
\Rightarrow \left\{ \begin{array}{l}
\widehat A = {5.12^0} = {60^0}\\
\widehat B = {6.12^0} = {72^0}\\
\widehat C = {4.12^0} = {48^0}
\end{array} \right.\\
\Rightarrow \widehat C < \widehat A < \widehat B\\
\Rightarrow AB < BC < AC\\
b)\\
\frac{{AD}}{{AB}} = \frac{{CD}}{{BC}}\\
\Rightarrow AD < CD\left( {do:AB < BC} \right)
\end{array}$
