Решебник по геометрии 8 класс Мерзляк Задание 500

\[Схематический\ рисунок.\]

\[Дано:\]

\[AM - биссектриса\ \angle A;\]

\[CK - биссектриса\ \angle C;\]

\[AB = BC = b;\]

\[AC = a.\]

\[Найти:\]

\[\text{MK.}\]

\[Решение.\]

\[1)\ \mathrm{\Delta}ABC - \ равнобедренный:\]

\[\angle A = \angle C;\]

\[AM = CK.\]

\[AM - биссектриса\ \angle A:\]

\[\frac{\text{AB}}{\text{BM}} = \frac{\text{AC}}{\text{CM}};\]

\[\frac{b}{\text{BM}} = \frac{a}{\text{CM}}\]

\[CM = \frac{a}{b} \bullet BM;\]

\[BC = CM + BM;\]

\[b = \frac{a}{b} \bullet BM + BM\]

\[b^{2} = a \bullet BM + b \bullet BM\]

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

\[CK - биссектриса\ \angle C:\]

\[\frac{\text{AC}}{\text{AK}} = \frac{\text{BC}}{\text{BK}}\]

\[\frac{a}{\text{AK}} = \frac{b}{\text{BK}};\]

\[AK = \frac{a}{b} \bullet BK;\]

\[AB = AK + BK.\]

\[b = \frac{a}{b} \bullet BK + BK\]

\[b^{2} = a \bullet BK + b \bullet BK\]

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

\[2)\ \mathrm{\Delta}KBM\sim\mathrm{\Delta}ABC - второй\ \]

\[признак:\]

\[\frac{\text{BK}}{\text{AB}} = \frac{\text{BM}}{\text{BC}} = \frac{b}{a + b};\]

\[\angle ABC = \angle KBM.\]

\[Отсюда:\]

\[\frac{\text{MK}}{\text{AC}} = \frac{\text{BK}}{\text{AB}} = \frac{b}{a + b}\]

\[MK = AC \bullet \frac{b}{a + b} = \frac{\text{ab}}{a + b}.\]

\[Ответ:\ \ \frac{\text{ab}}{a + b}.\]