\[\boxed{\mathbf{617}\mathbf{.}\mathbf{еуроки - ответы\ на\ пятёрку}}\]
\[Рисунок\ по\ условию\ задачи:\]
\[\mathbf{Дано:}\]
\[ABCD - трапеция;\]
\[BC = a;AD = b;\]
\[MN \parallel AD;\]
\[S_{\text{AMND}} = S_{\text{MBCN}}.\]
\[\mathbf{Найти:}\]
\[MN - ?\]
\[\mathbf{Решение.}\]
\[1)\ MN = x.\]
\[(b + x)BH = (a + b + 2x)\text{BF.}\]
\[4)\ S_{\text{ABCD}} = S_{\text{AMND}} + S_{\text{MBCN}}\]
\[aBH - xBH = aBF - bBF;\]
\[(a - x)BH = (a - b)\text{BF.}\]
\[5)\ \left\{ \begin{matrix} (b + x)BH = (a + b + 2x)\text{BF} \\ (a - x)BH = (a - b)\text{BF\ \ \ \ \ \ \ \ \ \ } \\ \end{matrix} \right.\ \]
\[\left\{ \begin{matrix} BH = \frac{a + b + 2x}{b + x} \bullet BF \\ BH = \frac{a - b}{a - x} \bullet BF\ \ \ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \]
\[6)\ (a - x)(a + b + 2x) =\]
\[= (b + x)(a - b)\]
\[a^{2} + ab + 2ax - ax - bx - 2x^{2} =\]
\[= ba - b^{2} + ax - bx\]
\[a^{2} + 2ax - ax - 2x^{2} + b^{2} - ax =\]
\[= 0\]
\[a^{2} - 2x^{2} + b^{2} = 0\]
\[2x^{2} = a^{2} + b^{2}\]
\[x^{2} = \frac{a^{2} + b^{2}}{2}\]
\[x = \frac{\sqrt{a^{2} + b^{2}}}{2}.\]
\[Ответ:MN = \frac{\sqrt{a^{2} + b^{2}}}{2}.\]