\[\boxed{\mathbf{309.ОК\ ГДЗ - домашка\ на}\ 5}\]
\[Рисунок\ по\ условию\ задачи:\]
\[\mathbf{Дано:}\]
\[\mathrm{\Delta}ABC;\]
\[\text{AB} \neq \text{BC} \neq \text{AC}\]
\[AH - высота;\]
\[AD - биссектриса.\]
\[\mathbf{Доказать:}\]
\[\angle HAD = \frac{\angle B - \angle C}{2}.\]
\[\mathbf{Доказательство.}\]
\[1)\ \angle BAD = \angle DAC\ \]
\[(так\ как\ AD - биссектриса).\]
\[2)\ Рассмотрим\ \mathrm{\Delta}HAD -\]
\[прямоугольный:\]
\[\angle HAD = 90{^\circ} - \angle HDA.\]
\[3)\ \angle HDA - внешний\ к\ \mathrm{\Delta}ADC:\]
\[\angle HDA = \angle C + \angle CAD.\]
\[4)\ \angle BAD = \angle DAC;\ \ \]
\[\angle HDA = \angle C + \angle CAD:\]
\[\angle HDA = \angle C + \angle BAD.\]
\[5)\ \angle HAD = 90{^\circ} - \angle HDA;\ \]
\[\angle HDA = \angle C + \angle BAD:\]
\[\ \angle HAD = 90{^\circ} - \angle C - \angle BAD.\]
\[6) + \ \frac{\angle HAD = 90{^\circ} - \angle C - \angle BAD}{\angle HAD = 90{^\circ} - \angle C - \angle CAD}\]
\[2\angle HAD =\]
\[= 180{^\circ} - (2\angle C + \angle BAD + \angle CAD).\]
\[\angle A + \angle B + \angle C = 180{^\circ}:\]
\[2\angle HAD =\]
\[= \angle A + \angle B + \angle C - 2\angle C - \angle A\]
\[2\angle HAD = \angle B - \angle C\]
\[\angle HAD = \frac{\angle B - \angle C}{2}\text{.\ }\]
\[Что\ и\ требовалось\ доказать.\]