Решебник по геометрии 8 класс Мерзляк Задание 483

\[Схематический\ рисунок.\]

\[Дано:\]

\[EHGF - квадрат;\]

\[BD - высота;\]

\[AC = a;\]

\[BD = h.\]

\[Найти:\]

\[\text{HE.}\]

\[Решение.\]

\[1)\ EHGF - квадрат:\]

\[EF \parallel HG;\]

\[\angle E = 90{^\circ};\ \ \]

\[HG = HE.\]

\[2)\ \mathrm{\Delta}AEH\sim\mathrm{\Delta}ADB - первый\ \]

\[признак:\]

\[\angle HAE = \angle BAD;\]

\[\angle AEH = \angle ADB.\]

\[Отсюда:\]

\[\frac{\text{HE}}{\text{BD}} = \frac{\text{AH}}{\text{AB}}\]

\[HE = \frac{BD \bullet AH}{\text{AB}}\]

\[HE = \frac{h \bullet AH}{\text{AB}}.\]

\[3)\ Для\ AC\ и\ HG\ и\ секущей\ AB:\]

\[\angle CAB = \angle GHB.\]

\[4)\ \mathrm{\Delta}HBG\sim\mathrm{\Delta}ABC - первый\ \]

\[признак:\]

\[\angle CAB = \angle GHB;\]

\[\angle ABC = \angle HBG.\]

\[Отсюда:\]

\[\frac{\text{HG}}{\text{AC}} = \frac{\text{HB}}{\text{AB}}\]

\[HG = \frac{AC \bullet HB}{\text{AB}}\]

\[HG = \frac{a \bullet HB}{\text{AB}}.\]

\[5)\ \frac{h \bullet AH}{\text{AB}} = \frac{a \bullet HB}{\text{AB}}\]

\[h \bullet AH = a \bullet HB\]

\[h \bullet (AB - HB) = a \bullet HB\]

\[h \bullet AB - h \bullet HB = a \bullet HB\]

\[h \bullet AB = (a + h) \bullet HB\]

\[\frac{\text{HB}}{\text{AB}} = \frac{h}{a + h}\text{\ \ \ }\]

\[HE = \frac{\text{ah}}{a + h}.\]

\[Ответ:\ \ \frac{\text{ah}}{a + h}.\]