一、动手操作如图,将正方形纸片ABCD折叠,使点B落在CD边上一点E(不与点C、D重合),AB的对应边为FE,且FE交边AD于点G,压平后得到折痕MN.二、探究发现:∠GBE的值不变,为45°证明由折叠知:BN=NE,∠ABC=∠FEN=∠A=∠C=90°.连接BE.设∠NBE=∠NEB=α.则∠ENC=2α,∠BEC=90°-α,∠FEB=90°-α.∴∠BEC=∠FEB=90°-α.过点B作BQ⊥FE交FE于点Q.在△BEC与△BEQ中,BE=BE,∠BEC=∠FEB, First, the hands-on operation shown in the figure, the square paper ABCD folded so that the point B falls on the edge of the CD E (not coincident with the point C, D), AB corresponding edge FE, and FE cross-edge AD at point G, After flattening get crease MN. Second, found that: ∠GBE value unchanged for the 45 ° proved by the fold that: BN = NE, ∠ ABC = ∠ FEN = ∠ A = ∠ C = 90 °. Let ∠ENC = 2α, ∠BEC = 90 ° -α, ∠FEB = 90 ° -α.∴∠BEC = ∠FEB = 90 ° -α. If point B is BQ⊥FE Make FE at point Q. In ΔBEC and ΔB BEQ, BE = BE, ∠BEC = ∠FEB,