\[Рисунок\ в\ учебнике.\]
\[1)\ Отношение\ S_{1}\ и\ S_{2}.\]
\[\textbf{а)}\ \frac{S_{1}}{S_{2}} = \frac{\frac{1}{2} \bullet 3 \bullet b \bullet \sin\alpha}{\frac{1}{2} \bullet b \bullet 1 \bullet \sin\alpha} = 3.\]
\[\textbf{б)}\ \frac{S_{1}}{S_{2}} = \frac{\frac{1}{2} \bullet 2 \bullet a \bullet \sin\alpha}{\frac{1}{2} \bullet a \bullet 4 \bullet \sin\alpha} = \frac{2}{4} = \frac{1}{2}.\]
\[\textbf{в)}\ \frac{S_{1}}{S_{2}} = \frac{\frac{1}{2} \bullet 1 \bullet 4 \bullet \sin\alpha}{\frac{1}{2} \bullet 2 \bullet 5 \bullet \sin\alpha} = \frac{4}{10} = \frac{2}{5}.\]
\[2)\ Докажем:\]
\[\frac{S_{1}}{S_{2}} = \frac{\frac{1}{2} \bullet AB \bullet AC \bullet \sin{\angle BAC}}{\frac{1}{2} \bullet A_{1}B_{1} \bullet A_{1}C_{1} \bullet \sin{\angle B_{1}A_{1}C_{1}}} =\]
\[= \frac{AB \bullet AC \bullet \sin{\angle BAC}}{A_{1}B_{1} \bullet A_{1}C_{1} \bullet \sin{\angle BAC}} =\]
\[= \frac{AB \bullet AC}{A_{1}B_{1} \bullet A_{1}C_{1}}.\]
\[Что\ и\ требовалось\ доказать.\]