\[\boxed{\mathbf{415.еуроки - ответы\ на\ пятёрку}}\]
\[Дано:\ \]
\[\angle hk\ и\ \angle hl - смежные;\ \]
\[\angle hk < \angle hl.\]
\[Доказать:\ \]
\[1)\ \angle hk = 90{^\circ} - \frac{1}{2}(\angle hl - \angle hk);\ \]
\[2)\ \angle hl = 90{^\circ} + \frac{1}{2}(\angle hl - \angle hk).\]
\[Доказательство.\]
\[1)\ \angle hk = 180{^\circ} - \angle\text{hl\ }\]
\[(как\ смежные).\]
\[\angle hk + \angle hk = 180{^\circ} - \angle hl + \angle hk\]
\[\left. \ 2\angle hk = 180{^\circ} - \angle hl + \angle hk\ \ \ \ \ \ \right|:2\]
\[\angle hk = 90{^\circ} - \frac{1}{2}(\angle hl - \angle hk).\]
\[2)\ \angle hl = 180{^\circ} - \angle\text{hk\ }\]
\[(как\ смежные).\]
\[\ \angle hl + \angle hl = 180{^\circ} - \angle hk + \angle hl\]
\[\left. \ 2\angle hl = 180{^\circ} - \angle hk + \angle hl\ \ \ \ \ \ \ \right|:2\]
\[\angle hl = 90{^\circ} - \frac{1}{2}(\angle hk - \angle hl) =\]
\[= 90{^\circ} + \frac{1}{2}(\angle hl - \angle hk).\]
\[Что\ и\ требовалось\ доказать.\]