ГДЗ по геометрии 9 класс Мерзляк Задание 97

\[Схематический\ рисунок.\]

\[Дано:\]

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

\[BD - биссектриса\ \angle B.\]

\[Доказать:\]

\[\frac{\text{AD}}{\text{CD}} = \frac{\sin{\angle C}}{\sin{\angle A}}.\]

\[Доказательство.\]

\[1)\ В\ \mathrm{\Delta}ABC:\]

\[\angle ABD = \angle CBD = \frac{1}{2}\angle B.\]

\[2)\ В\ \mathrm{\Delta}ABD:\]

\[\frac{\text{AD}}{\sin{\angle ABD}} = \frac{\text{BD}}{\sin{\angle A}}\]

\[BD = \frac{AD \bullet \sin{\angle A}}{\sin\left( \frac{1}{2}\angle B \right)}.\]

\[3)\ В\ \mathrm{\Delta}CBD:\]

\[\frac{\text{CD}}{\sin{\angle CBD}} = \frac{\text{BD}}{\sin{\angle C}}\]

\[BD = \frac{CD \bullet \sin{\angle C}}{\sin\left( \frac{1}{2}\angle B \right)};\]

\[AD \bullet \sin{\angle A} = CD \bullet \sin{\angle C};\]

\[AD\ :CD = \sin{\angle C}\ :\sin{\angle A}.\]

\[Что\ и\ требовалось\ доказать.\]