Решебник по геометрии 10 класс Атанасян ФГОС 844

\[\boxed{\mathbf{844.}ОК\ ГДЗ\ –\ домашка\ на\ 5}\]

\[Дано:\ \ \]

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

\[вписанная\ окружность\ (O,r)\]

\[касается\ \mathrm{\Delta}ABC\ в\ точках\]

\[M \in AC,N \in BC,L \in AB;\ \ \]

\[описанная\ окружность\ \]

\[радиуса\ \text{R.}\]

\[Доказать:\ \]

\[\frac{S_{\text{LMN}}}{S_{\text{ABC}}} = \frac{r}{2R}.\]

\[Доказательство.\]

\[1)\ OL = OM = ON = r;\ \ \]

\[\angle LOM = 180{^\circ} - \angle A:\]

\[\sin{\angle LOM} = \sin{\angle A};\]

\[\frac{S_{\text{LOM}}}{S_{\text{ABC}}} = \frac{\frac{1}{2} \bullet r \bullet r \bullet \sin{\angle LOM}}{\frac{1}{2} \bullet AB \bullet AC \bullet \sin{\angle A}} =\]

\[= \frac{r^{2}}{AB \bullet AC}.\]

\[Аналогично:\ \]

\[\frac{S_{\text{MON}}}{S_{\text{ABC}}} = \frac{r^{2}}{CA \bullet CB};\text{\ \ }\]

\[\frac{S_{\text{NOL}}}{S_{\text{ABC}}} = \frac{r^{2}}{BA \bullet BC}.\]

\[2)\ S_{\text{LMN}} = S_{\text{LOM}} + S_{\text{MON}} + S_{\text{NOL}} =\]

\[= \frac{r^{2}(AB + BC + CA)}{AB \bullet BC \bullet CA} \bullet S_{\text{ABC}}.\]

\[3)\ S_{\text{ABC}} = pr =\]

\[= \frac{AB + BC + CA}{2} \bullet r\ :\]

\[4)\ S_{\text{ABC}} = \frac{AB \bullet BC \bullet AC}{4R}:\]

\[\frac{S_{\text{LMN}}}{S_{\text{ABC}}} =\]

\[= \frac{2r}{AB \bullet BC \bullet CA} \bullet \frac{AB \bullet BC \bullet AC}{4R} =\]

\[= \frac{r}{2R}.\]

\[Что\ и\ требовалось\ доказать.\]