Connect CC' first, and we can get that CC' is the bisector of angle EC'D, so CB'=CD, AB'=AB, so B' is the diagonal midpoint, AC=2AB, so ∠ ACB = 30, and we can get the answer.

Solution: Connect CC ′, fold △ABE along AE to make point B ′ on AC, and fold △CEF along EF to make point C ′ on the intersection of EB ′ and AD. ∴EC = EC′，∴EC′c =∞。 ∴CB'=CD, and ∵AB'=AB, so B' is the midpoint of diagonal AC, that is, AC=2AB, so ∠ ACB = 30, ∴ COT ∠ ACB = COT30 = BC/AB = ∠ 3, BC.