Решебник по геометрии 9 класс Мерзляк Задание 282

\[Схематический\ рисунок.\]

\[Дано:\]

\[ABCD - прямоугольник;\]

\[AM - биссектриса\ \angle BAD;\]

\[BM = 10\ см;\]

\[MC = 14\ см.\]

\[Найти:\]

\[BE;\ DE.\]

\[Решение.\]

\[1)\ ABCD - прямоугольник:\]

\[\angle A = \angle B = 90{^\circ};\ \ \ \]

\[AD = BC;\]

\[BC = BM + MC = 24;\]

\[\angle BAM = \angle DAE = \frac{1}{2}\angle A = 45{^\circ}.\]

\[2)\ В\ \mathrm{\Delta}ABM:\]

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

\[\angle A = 45{^\circ};\]

\[\angle M = 180{^\circ} - \angle B - \angle A = 45{^\circ};\]

\[AB = BM = 10.\]

\[3)\ В\ \mathrm{\Delta}ABD:\]

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

\[BD = \sqrt{AB^{2} + AD^{2}};\]

\[BD = \sqrt{100 + 576} = \sqrt{676} = 26.\]

\[4)\ \mathrm{\Delta}BEM\sim\mathrm{\Delta}DEA - по\ двум\ углам:\]

\[\angle BME = \angle DAE;\]

\[\angle BEM = \angle DEA - вертикальные.\]

\[Отсюда:\]

\[\frac{\text{DE}}{\text{BE}} = \frac{\text{AD}}{\text{BM}} = \frac{24}{10}\text{\ \ }\]

\[DE = \frac{24}{10}BE;\]

\[BD = DE + BE = 26;\]

\[\frac{24}{10}BE + BE = 26\]

\[\frac{34}{10}BE = 26\ \]

\[BE = \frac{130}{17}.\]

\[DE = \frac{24}{10} \bullet \frac{130}{17} = \frac{312}{17}.\]

\[Ответ:\ \ \frac{130}{17}\ см;\ \frac{312}{17}\ см.\]