Решебник по геометрии 9 класс Мерзляк Задание 104

\[Схематический\ рисунок.\]

\[Дано:\]

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

\[AD - биссектриса\ \angle A.\]

\[Доказать:\]

\[\frac{\text{BD}}{\text{AB}} = \frac{\text{CD}}{\text{AC}}.\]

\[Доказательство.\]

\[1)\ В\ \mathrm{\Delta}ABC:\]

\[\angle BAD = \angle CAD = \frac{1}{2}\angle A.\]

\[2)\ В\ \mathrm{\Delta}ABD:\]

\[\angle CDA = 180{^\circ} - \angle BDA;\]

\[\sin{\angle CDA} = \sin{\angle BDA}.\]

\[\frac{\text{AB}}{\sin{\angle BDA}} = \frac{\text{BD}}{\sin{\angle BAD}}\]

\[\frac{\text{BD}}{\text{AB}} = \frac{\sin{\angle BAD}}{\sin{\angle BDA}} = \frac{\sin\left( \frac{1}{2}\angle A \right)}{\sin{\angle BDA}}.\]

\[3)\ В\ \mathrm{\Delta}ACD:\]

\[\frac{\text{AC}}{\sin{\angle CDA}} = \frac{\text{CD}}{\sin{\angle CAD}}\]

\[\frac{\text{CD}}{\text{AC}} = \frac{\sin{\angle CAD}}{\sin{\angle CDA}} = \frac{\sin\left( \frac{1}{2}\angle A \right)}{\sin{\angle BDA}}\]

\[CD\ :AC = BD\ :AB.\]

\[Что\ и\ требовалось\ доказать.\]