\[Рисунок\ в\ учебнике.\]
\[Дано:\ \ \]
\[\angle ABE = \angle EBF = \angle FBC;\]
\[BD - биссектриса\ \angle ABE;\]
\[BK - биссектриса\ \angle FBC;\]
\[\angle ABC = 90{^\circ}.\]
\[Найти:\]
\[\angle DBK.\]
\[Решение\]
\[1)\ \angle ABC:\]
\[\angle ABC = \angle ABE + \angle EBF + \angle FBC\]
\[90{^\circ} = \angle ABE + \angle ABE + \angle ABE\]
\[3\angle ABE = 90{^\circ}\]
\[\angle ABE = 30{^\circ}.\]
\[2)\ \angle ABE:\]
\[\angle ABE = \angle ABD + \angle DBE\]
\[30{^\circ} = \angle DBE + \angle DBE\]
\[2\angle DBE = 30{^\circ}\]
\[\angle DBE = 15{^\circ}.\]
\[3)\ \angle FBC:\]
\[\angle FBC = \angle FBK + \angle KBC\]
\[30{^\circ} = \angle FBK + \angle FBK\]
\[2\angle FBK = 30{^\circ}\]
\[\angle FBK = 15{^\circ}.\]
\[4)\ \angle DBK:\]
\[\angle DBK = \angle DBE + \angle EBF + \angle FBK\]
\[\angle DBK = 15{^\circ} + 30{^\circ} + 15{^\circ} = 60{^\circ}.\]
\[Ответ:\ \ 60{^\circ}.\]