ГДЗ по геометрии 8 класс Атанасян Задание 829

\[\boxed{\mathbf{829.ОК\ ГДЗ - домашка\ на}\ 5}\]

\[Рисунок\ по\ условию\ задачи:\]

\[\mathbf{Дано:}\]

\[ABCD - параллелограмм;\]

\[M \in AC;M \in PR;\]

\[PR \parallel AD;\ \]

\[M \in QT;QT \parallel AB;\]

\[P \in AB;Q \in BC;\]

\[R \in CD;T \in AD.\]

\[\mathbf{Доказать:}\]

\[S_{\text{MPBQ}} = S_{\text{MRDT}}.\]

\[\mathbf{Доказательство.}\]

\[1)\ Проведем\ перпендикуляры:\ \]

\[EF\bot AD;M \in EF;E \in BC;\]

\[F \in AD.\ \]

\[CH\bot AB;M \in GH;G \in AB;\]

\[H \in CD.\]

\[2)\ Обозначим:\]

\[PB = MQ = a;\ \]

\[PM = BQ = b;\]

\[TM = DR = c;\]

\[MR = TD = d;\]

\[ME = h_{1};\ \ MF = h_{2};\ \ \]

\[MG = g_{1};MH = g_{2}.\]

\[3)\ S_{\text{MPBQ}} = BQ \bullet ME = BP \bullet MG;\ \ \]

\[bh_{1} = ag_{1}.\]

\[S_{\text{MRDT}} = TD \bullet MF = DR \bullet MH;\ \ \ \]

\[dh_{2} = cg_{2}.\]

\[4)\ S_{\text{MPBQ}} \bullet S_{\text{MPBQ}} = bh_{1} \bullet ag_{1} =\]

\[= ah_{1} \bullet bg_{1} = S_{\text{MPBQ}} \bullet S_{\text{APMT}};\]

\[S_{\text{MPDT}} \bullet S_{\text{MPDT}} = dh_{2} \bullet cg_{2} =\]

\[= dg_{2} \bullet ch_{2} = S_{\text{MPBQ}} \bullet S_{\text{APMT}}.\]

\[Откуда:\]

\[S_{\text{MPBQ}}^{2} = S_{\text{MRDT}}^{2}\text{\ \ \ \ }\]

\[S_{\text{MPBQ}} = S_{\text{MRDT}}.\]

\[Что\ и\ требовалось\ доказать.\]