(1)解:在 $\triangle BDC$ 中,因?yàn)?$BD = BC,$$\angle ABC = 80^\circ,$所以 $\angle BDC = \angle BCD = \frac{180^\circ - \angle DBC}{2} = \frac{180^\circ - 80^\circ}{2} = 50^\circ。$在 $\triangle ABC$ 中,$\angle A = 40^\circ,$$\angle ABC = 80^\circ,$所以 $\angle ACB = 180^\circ - 40^\circ - 80^\circ = 60^\circ。$因?yàn)?$CE = BC,$所以 $\triangle BCE$ 中,$BC = CE,$$\angle BCE = 60^\circ,$因此 $\triangle BCE$ 是等邊三角形,所以 $\angle BEC = 60^\circ。$因?yàn)?$\angle BEC$ 是 $\triangle ABE$ 的外角,所以 $\angle ABE = \angle BEC - \angle A = 60^\circ - 40^\circ = 20^\circ。$
(2)解:$\angle BEC + \angle BDC = 110^\circ。$理由如下:設(shè) $\angle ABC = \alpha^\circ,$在 $\triangle BDC$ 中,$BD = BC,$所以 $\angle BDC = \frac{180^\circ - \alpha}{2}。$在 $\triangle ABC$ 中,$\angle ACB = 180^\circ - 40^\circ - \alpha = (140 - \alpha)^\circ。$在 $\triangle BEC$ 中,$CE = BC,$所以 $\angle BEC = \frac{180^\circ - \angle BCE}{2} = \frac{180^\circ - (140^\circ - \alpha)}{2} = \frac{40^\circ + \alpha}{2}。$因此 $\angle BEC + \angle BDC = \frac{40^\circ + \alpha}{2} + \frac{180^\circ - \alpha}{2} = \frac{220^\circ}{2} = 110^\circ。$