I happen to have = = 1, CCDCB CDDCD II, 1 1, and △AOD are all equal △COB (the answer is not unique) 12, (1) (2) (6); (3)(4)(5) (The answer is not unique) 13, AC⊥BC or AB=BC, etc. 14, 1; Root number 3-115,2; 1; 3 16，26 17，AC=BD 18，①③④ 19，420， 120ⅲ，2 1。 It is proved that the quadrilateral ABCD is a parallelogram, so AB is parallel to CD and AB=CD. Because BE⊥AC in E, DF⊥AC in F, ∠ AEB = ∠ CFD = 90. So △ABE is all equal to CDF, so AE=CF22. Prove: Because F and G are the midpoint of AB and AC respectively, FG is parallel to BC, and FG= 1/2BC. It can also be proved that EF is parallel to AC, and EF= 1/2AC. Because EF and GD are not parallel, the quadrilateral DEFG is trapezoidal. Because AD⊥BC, AG=CG, DG= 1/2AC, EF=DG, the trapezoid DEFG is an isosceles trapezoid.