ГДЗ по геометрии 8 класс Атанасян Задание 846

\[\boxed{\mathbf{846.ОК\ ГДЗ - домашка\ на}\ 5}\]

\[Рисунок\ по\ условию\ задачи:\]

\[\mathbf{Дано:}\]

\[\mathrm{\Delta}ABC;\ \]

\[\angle C = 90{^\circ};\]

\[S_{\text{OAB}} = S_{\text{OAC}} = S_{\text{OBC}}.\]

\[\mathbf{Доказать:}\]

\[OA² + OB² = 5OC².\]

\[\mathbf{Доказательство.}\]

\[1)\ Проведем\ перпендикуляры\ \]

\[из\ точки\ O\ на\ катеты.\ \]

\[2)\ Допустим:\]

\[OD = p;\ \ OE = q;\ \ AB = c;\ \ \]

\[AC = b;\ \ BC = a;\ \frac{1}{2}ab = S.\]

\[Получим:\]

\[S_{\text{AOB}} = S_{\text{OAC}} = S_{\text{OBC}} = \frac{S}{3}.\]

\[3)\ Запишем\ равенства:\]

1.

\(S_{\text{OAC}} = S_{\text{OBC}};\ \)

\[\frac{1}{2}bq = \frac{1}{2}ap = \frac{S}{3};\ \ \ \ \]

\[q = \frac{a}{3};\ \ \]

\[p = \frac{b}{3}.\]

\[2.\ \ \ \]

\[CD = q = \frac{a}{3};\ \ \ \ \ \]

\[BD = a - q = \frac{2q}{3};\ \ \]

\[EC = p = \frac{b}{3};\ \ \ \ \]

\[\ AE = b - p = \frac{2b}{3}.\]

\[3.\ \ \ p^{2} = OB^{2} - BD^{2} =\]

\[= OC^{2} - CD^{2};\ \ \]

\[\frac{b^{2}}{9} = OB^{2} - \frac{4a^{2}}{9} = OC^{2} - \frac{a^{2}}{9}.\]

\[4.\ \ \ \]

\[q^{2} = OC^{2} - CE^{2} = OA^{2} - AE^{2};\ \ \]

\[\frac{a^{2}}{9} = OC^{2} - \frac{b^{2}}{9} = OA^{2} - \frac{4b^{2}}{9}.\]

\[4)\ Получим:\]

\[OA^{2} = \frac{a^{2} + 4b^{2}}{9};\ \ \ \]

\[OB^{2} = \frac{4a^{2} + b^{2}}{9};\ \ \]

\[OC^{2} = \frac{a^{2} + b^{2}}{9}.\]

\[Откуда:\]

\[OA^{2} + OB^{2} =\]

\[= \frac{a^{2} + 4b^{2}}{9} + \frac{4a^{2} + b^{2}}{9} =\]

\[= \frac{5a^{2} + 5b^{2}}{9} = 5 \bullet \left( \frac{a^{2} + b^{2}}{9} \right) =\]

\[= 5OC^{2}.\]

\[\mathbf{Что\ и\ требовалось\ доказать.}\]