\[\boxed{\mathbf{324.ОК\ ГДЗ - домашка\ на}\ 5}\]
\[Дано:\ \]
\[\angle hk\ и\ \angle hl - смежные;\ \]
\[\angle hk < \angle hl.\]
\[Доказать:\ \]
\[\angle hk = 90{^\circ} - \frac{1}{2}(\angle hl - \angle hk);\ \]
\[\angle hl = 90{^\circ} + \frac{1}{2}(\angle hl - \angle hk).\]
\[Доказательство.\]
\[1)\ \angle hk = 180{^\circ} - \angle hl\ \]
\[(как\ смежные).\]
\[2)\ \angle hk + \angle hk =\]
\[= 180{^\circ} - \angle hl + \angle hk\]
\[\left. \ 2\angle hk = 180{^\circ} - \angle hl + \angle hk\ \ \ \ \right|:2\]
\[\angle hk = 90{^\circ} - \frac{1}{2}(\angle hl - \angle hk).\]
\[3)\ \angle hl = 180{^\circ} - \angle hk\ \]
\[(как\ смежные).\]
\[4)\ \angle hl + \angle hl =\]
\[= 180{^\circ} - \angle hk + \angle hl\]
\[\left. \ 2\angle hl = 180{^\circ} - \angle hk + \angle hl\ \ \ \right|:2\]
\[\angle hl = 90{^\circ} - \frac{1}{2}(\angle hk - \angle hl) =\]
\[= 90{^\circ} + \frac{1}{2}(\angle hl - \angle hk).\]
\[Что\ и\ требовалось\ доказать.\]