\[Дано:\]
\[\mathrm{\Delta}ABC;\]
\[AC^{2} = AB^{2} + AB \bullet BC + BC^{2}.\]
\[Доказать:\]
\[\angle B = 120{^\circ}.\]
\[Доказательство.\]
\[AC^{2} = AB^{2} + BC^{2} - 2AB \bullet BC\cos{\angle B}\]
\[AB \bullet BC = - 2AB \bullet BC \bullet \cos{\angle B}\]
\[\cos{\angle B} = - \frac{1}{2}\text{\ \ \ }\]
\[\angle B = 120{^\circ}.\]
\[Что\ и\ требовалось\ доказать.\]