\[Схематический\ рисунок.\]
\[Дано:\]
\[BD - биссектриса\ \angle B;\]
\[\angle ABC = 80{^\circ}.\]
\[Найти:\]
\[углы\ \mathrm{\Delta}ADC.\]
\[Решение.\]
\[\cup ADC = 2\angle ABC = 160{^\circ};\]
\[\cup ADC + \cup ABC = 360{^\circ}\]
\[160{^\circ} + \cup ABC = 360{^\circ}\]
\[\cup ABC = 200{^\circ}.\]
\[\angle ADC = \frac{1}{2} \cup ABC = 100{^\circ};\]
\[\angle ABD = \frac{1}{2}\angle ABC = 40{^\circ};\]
\[\angle CBD = \angle ABD = 40{^\circ};\]
\[\angle ACD = \angle ABD = 40{^\circ};\]
\[\angle CAD = \angle CBD = 40{^\circ}.\]
\[Ответ:\ \ 40{^\circ};\ 100{^\circ};\ 40{^\circ}.\]