ГДЗ по геометрии 11 класс. Атанасян ФГОС 846

\[\boxed{\mathbf{846.}ОК\ ГДЗ\ –\ домашка\ на\ 5}\]

\[Дано:\ \ \]

\[ABCD - четырехугольник;\ \ \]

\[AB = b;\ \ BC = c;\ \ CD = d;\ \ \]

\[DA = a;\]

\[p - полуперимтер;\ \ \]

\[вневписанные\ окружности\ \]

\[\left( O_{1};r_{a} \right);\ \ \left( O_{2};r_{c} \right)\text{.\ \ \ }\]

\[Доказать:\ \text{\ \ }\]

\[S_{\text{ABCD}} = r_{a}(p - a) + r_{c}(p - c).\]

\[Доказательство.\]

\[1)\ Касательные\ из\ одной\ \]

\[точки:\ \]

\[AT = AM,\ \ \ DT = DN,\ \ \ BQ = BK,\ \ \ \]

\[CL = CQ.\]

\[2)\ Пусть\ AT = AM = k;\ \ \]

\[DT = DN = a - k;\ \ \]

\[BQ = BK = l:\]

\[CL = CQ = c - l.\]

\[l + b + l = a - k + d + c - l\ \ \]

\[2k + 2l = (a - b + c + d)\]

\[k + l = \frac{a - b + c + d}{2} = p - b.\]

\[Отсюда:\]

\[KM = k + b + l = p - b + b = p.\]

\[4)\ \ O_{1}M\bot MK\ и\ \ O_{2}K\bot MK:\ \]

\[O_{1}M \parallel O_{2}\text{K.}\]

\[Аналогично:\ \ \ O_{1}N \parallel O_{2}\text{L.}\]

\[Значит:\ \]

\[O_{1}\text{MK}O_{2}\ и\ \ O_{1}O_{2}LN -\]

\[прямоугольные\ трапеции.\]

\[S_{O_{1}\text{MK}O_{2}\text{LN}} = S_{O_{1}\text{MK}O_{2}} + S_{O_{1}O_{2}\text{LN}} =\]

\[= \frac{1}{2}\text{MK}\left( r_{a} + r_{c} \right) + \frac{1}{2}\text{LN}\left( r_{a} + r_{c} \right) =\]

\[= MK\left( r_{a} + r_{c} \right) = p\left( r_{a} + r_{c} \right).\]

\[5)\ S_{O_{1}\text{MADN}} = 2S_{O_{1}\text{AT}} + 2S_{O_{1}\text{TD}} =\]

\[= r_{a}k + r_{a}(a - k) =\]

\[= r_{a}k + r_{a}a - r_{a}k = r_{a}a;\]

\[S_{\text{BK}O_{2}\text{LC}} = 2S_{\text{KBQ}} + 2S_{\text{LCQ}} =\]

\[= r_{c}l + r_{c}(c - l) =\]

\[= r_{c}l + r_{c}c - r_{c}l = r_{c}\text{c.}\]

\[6)\ Таким\ образом:\]

\[S_{\text{ABCD}} =\]

\[= S_{O_{1}\text{MK}O_{2}\text{LN}} - S_{O_{1}\text{MADN}} - S_{\text{BK}O_{2}\text{LC}} =\]

\[= p\left( r_{a} + r_{c} \right) - r_{a}a - r_{c}c =\]

\[= pr_{a} + pr_{c} - r_{a}a - r_{c}c =\]

\[= r_{a}(p - a) + r_{a}(p - c).\]

\[Что\ и\ требовалось\ доказать.\]