\[Рисунок\ в\ учебнике.\]
\[Дано:\]
\[AB \parallel CD;\]
\[AB = AE;\]
\[CD = CE.\]
\[Доказать:\]
\[BE\bot DE.\]
\[Доказательство.\]
\[1)\ \mathrm{\Delta}ABE - \ равнобедренный:\]
\[\angle ABE = \angle AEB;\]
\[\angle ABE + \angle AEB + \angle BAE = 180{^\circ}\]
\[\angle BAE = 180{^\circ} - 2\angle AEB.\]
\[2)\ \mathrm{\Delta}CDE - \ равнобедренный:\]
\[\angle CDE = \angle CED;\]
\[\angle CDE + \angle CED + \angle DCE = 180{^\circ}\]
\[\angle DCE = 180{^\circ} - 2\angle CED.\]
\[3)\ Для\ прямых\ \text{AB\ }и\ \text{CD\ }и\ \]
\[секущей\ AC:\]
\[\angle BAC + \angle DCE = 180{^\circ}\]
\[180{^\circ} - 2\angle AEB + 180{^\circ} - 2\angle CED = 180{^\circ}\]
\[2\angle AEB + 2\angle CED = 180{^\circ}\]
\[\angle AEB + \angle CED = 90{^\circ}.\]
\[4)\ \angle AEC = \angle AEB + \angle BED + \angle CED\]
\[180{^\circ} = 90{^\circ} + \angle BED\ \ \ \]
\[\angle BED = 90{^\circ}.\]
\[Что\ и\ требовалось\ доказать.\]