\[\boxed{\mathbf{958.ОК\ ГДЗ - домашка\ на}\ 5}\]
\[Рисунок\ по\ условию\ задачи:\]
\[\mathbf{Дано:}\]
\[ABCD - прямоугольник.\]
\[\mathbf{Доказать:}\]
\[для\ любой\ точки\ M\]
\[AM^{2} + CM^{2} = BM^{2} + DM^{2}.\]
\[\mathbf{Доказательство.}\]
\[1)\ Введем\ прямоугольную\ \]
\[систему\ координат:\ \]
\[A(0;0);\ D(a;0);B(0;c);C(a;c);\]
\[M(x;y).\]
\[2)\ AM = \sqrt{(x - 0)^{2} + (y - 0)^{2}} =\]
\[= \sqrt{x^{2} + y^{2}}\]
\[AM^{2} = x^{2} + y^{2}.\]
\[CM = \sqrt{(x - a)^{2} + (y - c)^{2}}\]
\[CM^{2} = (x - a)^{2} + (y - c)^{2}.\]
\[BM = \sqrt{(x - 0)^{2} + (y - c)^{2}} =\]
\[= \sqrt{x^{2} + {(y - c)}^{2}}\]
\[BM^{2} = x^{2} + {(y - c)}^{2}.\]
\[DM = \sqrt{(x - a)^{2} + (y - 0)^{2}} =\]
\[= \sqrt{{(x - a)}^{2} + y^{2}}\]
\[DM^{2} = {(x - a)}^{2} + y^{2}.\]
\[3)\ AM^{2} + CM^{2} =\]
\[= x^{2} + y^{2} + (x - a)^{2} + (y - c)^{2};\]
\(BM^{2} + DM^{2} =\)
\[= x^{2} + (y - c)^{2} + (x - a)^{2} + y^{2};\ \]
\[\ AM^{2} + CM^{2} = BM^{2} + DM^{2}.\]
\[Что\ и\ требовалось\ доказать.\]
\[\mathbf{Параграф}\ 3\mathbf{.\ Уравнения\ окружности\ и\ прямой}\]