解：在$\triangle ABC$中，$\angle BAC + \angle B + \angle ACB = 180^{\circ}。$
因为$\angle ACB = 40^{\circ}，$所以$\angle BAC + \angle B = 140^{\circ}。$
又因为$\angle BAC = 6\angle B，$所以$6\angle B + \angle B = 140^{\circ}，$
即$7\angle B = 140^{\circ}，$解得$\angle B = 20^{\circ}。$
因为$\angle ACB = 40^{\circ}，$所以$\angle ACE = 180^{\circ}-\angle ACB = 140^{\circ}。$
因为$CD$平分$\angle ACE，$所以$\angle DCE=\frac{1}{2}\angle ACE=\frac{1}{2}\times140^{\circ}=70^{\circ}。$
所以$\angle D = \angle DCE - \angle B = 70^{\circ}-20^{\circ}=50^{\circ}。$
