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1、.............o............o............o.............o.............外.............o............o.............o.............装.............o.............o............o.............订.............o.............o.............o.............线.............o.............o
2、..........................o.............o............o.........学校：___________姓名：____________班级：____________考号：____________.............o.............o..........................o.............内.............o.............o.............o...........装.............o...
3、..........o.............o............订.............o.............o.............o............线.............o.............o.............o.............o.............o.............o........绝密★启用前乐学教育菱形证明专题训练1.已知:如图,在四边形ABCD中,AB∥CD,E,F为对角线AC上两点,且AE=CF,DF∥BE,AC平分∠BAD.
4、求证:四边形ABCD为菱形. 【答案】∵AB∥CD, ∴∠BAE=∠DCF. ∵DF∥BE, ∴∠BEF=∠DFE, ∴∠AEB=∠CFD. 又∵AE=CF, ∴△AEB≌∠CFD, ∴AB=CD. ∵AB∥CD, ∴四边形ABCD是平行四边形. ∵AC平分∠BAD, ∴∠BAE=∠DAF. 又∠BAE=∠DCF, ∴∠DAF=∠DCF, ∴AD=CD, ∴四边形ABCD是菱形.2.如图，矩形ABCD中，点O为AC的中点，过点O的直线分别与AB，CD交于点E，F，连接BF交AC于点M，连接DE，BO.若∠COB=60°
5、，FO=FC. 求证:(1)四边形EBFD是菱形;【答案】连接OD.∵点O为矩形ABCD的对角线AC的中点， ∴B,D, O三点共线且BD=DO=CO=AO. 在矩形ABCD中，AB∥DC，AB=DC，∴∠FCO=∠EAO. 在△CFO和△AEO中， ∴△CFO≌△AEO，∴FO=EO. 又∵BO=DO，∴四边形BEFD是平行四边形. ∵BO=CO，∠COB=60°， ∴△COB是等边三角形.∴∠OCB=60°.第19页共19页.............o............o............o.....
6、........o.............外.............o............o.............o.............装.............o.............o............o.............订.............o.............o.............o.............线.............o.............o..........................o.............o....
7、........o.........学校：___________姓名：____________班级：____________考号：____________.............o.............o..........................o.............内.............o.............o.............o...........装.............o.............o.............o............订.......
8、......o.............o.............o............线.............o.............o.............o.............o.............o.............o........ ∴∠FCO=∠DCB-∠OCB=30°. ∵FO=
