\[\boxed{\mathbf{1273.ОК\ ГДЗ - домашка\ н}а\ 5}\]
\[Рисунок\ по\ условию\ задачи:\]
\[Дано:\]
\[окружность\ O(0,\ R);\]
\[ABCD - вписанный\ \]
\[четырехугольник;\]
\[AB = a;\ \]
\[BC = b;\ \]
\[CD = c;\ \]
\[AD = d.\]
\[Найти:\]
\[AC,\ BD - ?\]
\[Решение.\]
\[1)\ По\ свойству\ вписанного\ \]
\[четырехугольника:\ \]
\[\angle A + \angle C = 180{^\circ};\]
\[\angle B + \angle D = 180{^\circ}.\]
\[2)\ По\ теореме\ косинусов:\]
\[AC^{2} = a^{2} + b^{2} - 2ab\cos B;\]
\[cosB = \frac{a^{2} + b^{2} - c^{2} - d^{2}}{2(ab + cd)}.\]
\[AC^{2} =\]
\[= a^{2} + b^{2} - 2ab \cdot \frac{a^{2} + b^{2} - c^{2} - d^{2}}{2(ab + cd)} =\]
\[= \frac{\text{cd}\left( a^{2} + b^{2} \right) + ab(c^{2} + d^{2})}{ab + cd}.\]
\[AC = \sqrt{\frac{\text{cd}\left( a^{2} + b^{2} \right) + ab(c^{2} + d^{2})}{ab + cd}}.\]
\[3)\ Аналогично\ для\ второй\ \]
\[диагонали:\]
\[BD^{2} = a^{2} + d^{2} - 2ad\cos A;\]
\[BD^{2} = b^{2} + c^{2} - 2b\ cosA =\]
\[= b^{2} + c^{2} + 2bc\cos A.\]
\[a^{2} + d^{2} - 2ad\ cosA =\]
\[= b^{2} + c^{2} + 2bc\cos A\]
\[2(ad + bc)cosA =\]
\[= a^{2} + d^{2} - b^{2} - c^{2}\]
\[socA = \frac{a^{2} + d^{2} - b^{2} - c^{2}}{2(ad + bc)}.\]
\[BD^{2} =\]
\[= a^{2} + d^{2} - 2ad \cdot \frac{a^{2} + d^{2} - b^{2} - c^{2}}{2(ad + bc)} =\]
\[= \frac{\text{bc}\left( a^{2} + d^{2} \right) + ad(b^{2} + c^{2})}{ad + bc}.\]
\[BD =\]
\[= \sqrt{\frac{\text{bc}\left( a^{2} + d^{2} \right) + ad(b^{2} + c^{2})}{ad + bc}}.\]
\[Ответ:\]
\[\sqrt{\frac{\text{cd}\left( a^{2} + b^{2} \right) + ab\left( c^{2} + d^{2} \right)}{ab + cd}};\ \]
\[\sqrt{\frac{\text{bc}\left( a^{2} + d^{2} \right) + ad(b^{2} + c^{2})}{ad + bc}}.\]