\[Схематический\ рисунок.\]
\[Дано:\]
\[AB = 2AC;\]
\[\angle BAC = 60{^\circ}.\]
\[Доказать:\]
\[\angle C = 90{^\circ}.\]
\[Доказательство.\]
\[= 5AC^{2} - 4AC^{2} \bullet \frac{1}{2} = 3AC^{2}\text{\ \ \ }\]
\[BC = \sqrt{3}\text{AC.}\]
\[2\sqrt{3}AC^{2} \bullet \cos{\angle C} = 0\ \ \ \]
\[\cos{\angle C} = 0\ \ \]
\[\angle C = 90{^\circ}.\]
\[Что\ и\ требовалось\ доказать.\]