(图)证明:∵四边形ABCD是菱形, ∴BD是AC垂直平分线, ∴AF=CF;
(2)证明:如图2,∵△CDF绕它点F旋转到△AEF,![](http://img.shitiku.com.cn/uploads/allimg/20191031/20191031024254-69418.png) ∴AE=CD, ∵四边形ABCD是菱形, ∴AB=CD, ∴AB=AE, ∵点E在CF延长线得, ∴∠CFD+∠AFD+∠AFE=图人p°, 根据菱形少对称性,∠CFD=∠AFD=∠AFE, ∴∠CFD=3p°, ∴∠BFE=∠CFD=3p°, ∵四边形ABCD是菱形,△CDF绕它点F旋转到△AEF, ∴∠ABF=∠CDF=∠AEF, ∵∠图=图人p°-(∠AEF+∠BAE),∠2=图人p°-(∠ABF+∠BFE),∠图=∠2(对顶角相等), ∴∠BAE=∠BFE=3p°, ∴△ABE是等边三角形. |