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МатематикаГДЗ по геометрии 11 класс. Атанасян ФГОС 845
845
\[\boxed{\mathbf{845.}ОК\ ГДЗ\ –\ домашка\ на\ 5}\]
\[Дано:\ \ \]
\[\mathrm{\Delta}ABC;\ \ \ \]
\[вневписанные\ окружности\ \]
\[радиусом\ r_{a},r_{b},r_{c};\]
\[вписанная\ окружность\ \]
\[радиусом\ r;\ \]
\[BC = a.\]
\[Доказать:\ \text{\ \ }\]
\[\textbf{а)}\ S_{\text{ABC}} = r_{a}(p - a);\ \ \]
\[\textbf{б)}\ \ S = \sqrt{r \bullet r_{a} \bullet r_{b} \bullet r_{c}};\ \ \]
\[\frac{1}{r_{a}} + \frac{1}{r_{b}} + \frac{1}{r_{c}} = \frac{1}{r}.\]
\[Доказательство.\]
\[\textbf{а)}\ Пусть\ O - центр\ \]
\[вневписанной\ окружности;\ \]
\[AB = c,\ BC = a,\ AC = b.\]
\[1)\ AP = AT,\ CH = CT,\]
\[BH = BP = k\ \]
\[(как\ касательные\ из\ одной\ точки):\]
\[\ AP = c + k;\ \ \]
\[AT = AC + BC - BH =\]
\[= b + a - k;\]
\[b + a - k = c + k\]
\[k = \frac{b + a - c}{2} = p - c;\ \]
\[где\ p - полупериметр.\]
\[AT = AP = c + k = c + p - c =\]
\[= p;\ \ \ \]
\[CH = CT = a - k.\]
\[2)\ S_{\text{APOT}} = S_{\text{AOP}} + S_{\text{AOT}} =\]
\[= \frac{1}{2}r_{a} \bullet p + \frac{1}{2}r_{a} \bullet p = r_{a}\text{p.}\]
\[3)\ S_{\text{OPBCT}} =\]
\[= S_{\text{OPB}} + S_{\text{BOH}} + S_{\text{OHC}} + S_{\text{OCT}} =\]
\[= r_{a} \bullet k + r_{a}(a - k) = r_{a} \bullet a.\]
\[4)\ Таким\ образом:\]
\[S_{\text{ABC}} = S_{\text{APOT}} - S_{\text{OPBCT}} =\]
\[= r_{a}p - r_{a}a = r_{a}(p - a).\]
\[Что\ и\ требовалось\ доказать.\]
\[\textbf{б)}\ S_{\text{ABC}} = r_{a}(p - a) = pr:\ \]
\[\frac{S_{\text{ABC}}}{r} = p;\]
\[\ \frac{S_{\text{ABC}}}{r_{a}} = p - a.\]
\[Аналогично:\ \ \]
\[\frac{S_{\text{ABC}}}{r_{b}} = p - b\ \ \ и\ \ \frac{S_{\text{ABC}}}{r_{c}} = p - c.\]
\[1)\ S_{\text{ABC}} =\]
\[= \sqrt{p(p - a)(p - b)(p - c)} =\]
\[= \sqrt{\frac{S_{\text{ABC}}}{r} \bullet \frac{S_{\text{ABC}}}{r_{a}} \bullet \frac{S_{\text{ABC}}}{r_{b}} \bullet \frac{S_{\text{ABC}}}{r_{c}}} =\]
\[= \frac{S_{\text{ABC}}^{2}}{\sqrt{rr_{a}r_{b}r_{c}}}\]
\[S_{\text{ABC}} = \sqrt{rr_{a}r_{b}r_{c}}.\]
\[2)\ S_{\text{ABC}} = r_{a}(p - a) = pr:\]
\[\frac{1}{3} = \frac{p - a}{S_{\text{ABC}}};\]
\[\ \frac{1}{r_{a}} = \frac{p - a}{S_{\text{ABC}}}.\]
\[Аналогично:\ \ \]
\[\frac{1}{r_{b}} = \frac{p - b}{S_{\text{ABC}}};\text{\ \ }\]
\[\frac{1}{r_{c}} = \frac{p - c}{S_{\text{ABC}}}.\]
\[3)\ \frac{1}{r_{a}} + \frac{1}{r_{b}} + \frac{1}{r_{c}} =\]
\[= \frac{(p - a) + (p - b) + (p - c)}{S_{\text{ABC}}} =\]
\[= \frac{3p - (a + b + c)}{S_{\text{ABC}}} =\]
\[= \frac{3p - 2p}{S_{\text{ABC}}} = \frac{p}{S_{\text{ABC}}} = \frac{1}{r}.\]
\[Что\ и\ требовалось\ доказать.\]
