如图,在平行四边形ABCD中,AE⊥BC,垂足为点E,CE=CD,F为CE中点,G为CD上一点,连接DF,EG,AG,且∠1=∠2.(1)若CF=2,AE=3,求BE长; (2)求证:∠CEG=1/2∠AGE.
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![如图,在平行四边形ABCD中,AE⊥BC,垂足为点E,CE=CD,F为CE中点,G为CD上一点,连接DF,EG,AG,且∠1=∠2.(1)若CF=2,AE=3,求BE长; (2)求证:∠CEG=1/2∠AGE.](/uploads/image/z/2394813-21-3.jpg?t=%E5%A6%82%E5%9B%BE%2C%E5%9C%A8%E5%B9%B3%E8%A1%8C%E5%9B%9B%E8%BE%B9%E5%BD%A2ABCD%E4%B8%AD%2CAE%E2%8A%A5BC%2C%E5%9E%82%E8%B6%B3%E4%B8%BA%E7%82%B9E%2CCE%3DCD%2CF%E4%B8%BACE%E4%B8%AD%E7%82%B9%2CG%E4%B8%BACD%E4%B8%8A%E4%B8%80%E7%82%B9%2C%E8%BF%9E%E6%8E%A5DF%2CEG%2CAG%2C%E4%B8%94%E2%88%A01%3D%E2%88%A02.%EF%BC%881%EF%BC%89%E8%8B%A5CF%3D2%2CAE%3D3%2C%E6%B1%82BE%E9%95%BF%EF%BC%9B+++++++++++++++++++++++++++++++%EF%BC%882%EF%BC%89%E6%B1%82%E8%AF%81%EF%BC%9A%E2%88%A0CEG%3D1%EF%BC%8F2%E2%88%A0AGE.)
如图,在平行四边形ABCD中,AE⊥BC,垂足为点E,CE=CD,F为CE中点,G为CD上一点,连接DF,EG,AG,且∠1=∠2.(1)若CF=2,AE=3,求BE长; (2)求证:∠CEG=1/2∠AGE.
如图,在平行四边形ABCD中,AE⊥BC,垂足为点E,CE=CD,F为CE中点,G为CD上一点,连接DF,EG,AG,且∠1
=∠2.
(1)若CF=2,AE=3,求BE长;
(2)求证:∠CEG=1/2∠AGE.
如图,在平行四边形ABCD中,AE⊥BC,垂足为点E,CE=CD,F为CE中点,G为CD上一点,连接DF,EG,AG,且∠1=∠2.(1)若CF=2,AE=3,求BE长; (2)求证:∠CEG=1/2∠AGE.
∵CE=CD,点F为CE的中点,CF=2,
∴DC=CE=2CF=4,
∵四边形ABCD是平行四边形,
∴AB=CD=4,
∵AE⊥BC,
∴∠AEB=90°,
在Rt△ABE中,由勾股定理得:BE=根号4²-3²=根号7
(2)
过G作GM⊥AE于M,
∵AE⊥BE,
∴GM∥BC∥AD,
∵在△DCF和△ECG中,
∠1=∠2
∠C=∠C
CD=CE
∴△DCF≌△ECG(AAS),
∴CG=CF,
∵CE=CD,CE=2CF,
∴CD=2CG
即G为CD中点,
∵AD∥GM∥BC,
∴M为AE中点,
∵GM⊥AE,
∴AM=EM,
∴∠AGE=2∠MGE,
∵GM∥BC,
∴∠EGM=∠CEG,
∴∠CEG=1/2∠AGE.