Решебник по геометрии 7 класс Мерзляк Задание 418

\[\boxed{\mathbf{418}.еуроки - ответы\ на\ пятёрку}\]

\[Рисунок\ по\ условию\ задачи:\]

\[Дано:\]

\[\mathrm{\Delta}ABC;\]

\[AB = BC;\]

\[E \in AB;\]

\[F \in BC;\]

\[AC = AF = EF = BE.\]

\[Найти:\]

\[\angle A;\ \angle B;\ \angle C.\]

\[Решение.\]

\[1)\ \mathrm{\Delta}ABC - равнобедренный:\]

\[\angle A = \angle C = x;\]

\[\angle A + \angle B + \angle C = 180{^\circ}\]

\[\angle B = 180{^\circ} - \angle A - \angle C =\]

\[= 180{^\circ} - x - x = 180{^\circ} - 2x.\]

\[2)\ \mathrm{\Delta}FAC - равнобедренный\ \]

\[(AC = AF):\]

\[\angle C = \angle AFC = x;\]

\[\angle C + \angle AFC + \angle FAC = 180{^\circ}\]

\[\angle FAC = 180 - x - x =\]

\[= 180{^\circ} - 2x.\]

\[3)\ \angle A = x = \angle EAF + \angle FAC\]

\[x = \angle EAF + 180 - 2x\]

\[\angle EAF = x + 2x - 180 =\]

\[= 3x - 180{^\circ}.\]

\[\angle EAF + \angle AEF + \angle EFA = 180{^\circ}\]

\[\angle EFA =\]

\[= 180 - (3x - 180 + 3x - 180) =\]

\[= 540{^\circ} - 6x.\]

\[5)\ \mathrm{\Delta}BEF - равнобедренный\ \]

\[(BE = EF):\]

\[\angle EBF = \angle BFE = 180{^\circ} - 2x.\]

\[6)\ \angle CFB = 180{^\circ} -\]

\[развернутый.\]

\[\angle CFB =\]

\[= \angle AFC + \angle AFE + \angle EFB =\]

\[= 180{^\circ}\]

\[x + 540 - 6x + 180 - 2x = 180\]

\[7x = 540{^\circ}\]

\[x = \frac{540}{7} = 77\frac{1}{7}{^\circ}.\]

\[\angle A = \angle C = 77\frac{1}{7}{^\circ}.\]

\[7)\ \angle B = 180 - 2 \cdot \frac{540}{7} = 25\frac{5}{7}{^\circ}.\]

\[Ответ:\ \angle A = \angle C = 77\frac{1}{7}{^\circ};\ \ \]

\[\angle B = 25\frac{5}{7}{^\circ}.\]