\[\boxed{\mathbf{509}\mathbf{.}\mathbf{ОК}\mathbf{\ }\mathbf{ГДЗ}\mathbf{-}\mathbf{домашка}\mathbf{\ }\mathbf{на}\ 5}\]
\[Рисунок\ по\ условию\ задачи:\]
\[\mathbf{Дано:}\]
\[\mathrm{\Delta}\text{ABC} - равносторонний;\]
\[\text{AB} = \text{BC} = \text{AC};\]
\(\text{EN}\bot\text{BC};\text{OH}\bot\text{AC};\)
\[\text{EK}\bot\text{AC};\text{OF}\bot\text{AB};\ \]
\[\text{EM}\bot\text{AB};\text{OQ}\bot\text{BC}.\]
\[\mathbf{Доказать:}\]
\[\text{EN} + \text{EK} + \text{EM} =\]
\[= \text{OQ} + \text{OH} + \text{OF}.\]
\[\mathbf{Доказательство.}\]
\[1)\ \left. \ \frac{S_{\text{ABC}} = S_{\text{AEC}} + S_{\text{BEC}} + S_{\text{ABE}}}{S_{\text{ABC}} = S_{\text{AOB}} + S_{\text{BOC}} + S_{\text{AOC}}} \right| \Longrightarrow\]
\[\Longrightarrow S_{\text{AEC}} + S_{\text{BEC}} + S_{\text{ABE}} =\]
\[= S_{\text{AOB}} + S_{\text{BOC}} + S_{\text{AOC}}.\]
\[\text{AB} = \text{BC} = \text{AC}\ (по\ условию);\]
\[\frac{1}{2}\text{AC}\left( \text{OH} + \text{OQ} + \text{OF} \right) =\]
\[= \frac{1}{2}\text{AC}\left( \text{EK} + \text{EN} + \text{EM} \right);\]
\[\text{EN} + \text{EK} + \text{EM} =\]
\[= \text{OQ} + \text{OH} + \text{OF}.\ \]
\[Что\ и\ требовалось\ доказать.\]
\[\boxed{\mathbf{509.еуроки - ответы\ на\ пятёрку}}\]
\[Рисунок\ по\ условию\ задачи:\]
\[\mathbf{Дано:}\]
\[ABCD - ромб;\]
\[\angle A = 45{^\circ}.\]
\[\mathbf{Найти:}\]
\[\angle\text{ABD};\ \angle\text{BAC.}\]
\[\mathbf{Решение.}\]
\[1)\ По\ свойству\ ромба:\]
\[\angle A = \angle C;\]
\[\angle B = \angle D.\]
\[2)\ \angle A + \angle B + \angle C + \angle D = 360{^\circ}\]
\[2\angle A + 2\angle B = 360{^\circ}\]
\[2\angle B = 360{^\circ} - 90{^\circ} = 270{^\circ}\]
\[\angle B = \angle D = 135{^\circ}.\]
\[3)\ BD - биссектриса\ \angle B:\]
\[\angle ABD = \frac{1}{2}\angle B = 67{^\circ}30^{'}\]
\[(по\ свойству\ ромба).\]
\[4)\ AC - биссектриса\ \angle A:\]
\[\angle BAC = \frac{1}{2}\angle A = 22{^\circ}30^{'}\]
\[(по\ свойству\ ромба).\]
\[Ответ:\ \angle ABD = 67{^\circ}30^{'};\ \]
\[\angle BAC = 22{^\circ}30^{'}.\]