Known: as shown in the figure, CE bisects ∠ ACD, ∠ 1 = ∠ B, verification: ab ‖ CE

Known: as shown in the figure, CE bisects ∠ ACD, ∠ 1 = ∠ B, verification: ab ‖ CE

It is proved that ∵ CE bisects ∵ ACD, ∵ 1 = ∵ 2, ∵ 1 = ∵ B, ∵ 2 = ∵ B, ∥ ab ∥ CE