如图E为菱形ABCD边BC上一点且AB=AE AE交BD于O 且∠DAE=2∠BAE 求证:EB=OA
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![如图E为菱形ABCD边BC上一点且AB=AE AE交BD于O 且∠DAE=2∠BAE 求证:EB=OA](/uploads/image/z/1262047-31-7.jpg?t=%E5%A6%82%E5%9B%BEE%E4%B8%BA%E8%8F%B1%E5%BD%A2ABCD%E8%BE%B9BC%E4%B8%8A%E4%B8%80%E7%82%B9%E4%B8%94AB%3DAE+AE%E4%BA%A4BD%E4%BA%8EO+%E4%B8%94%E2%88%A0DAE%3D2%E2%88%A0BAE+%E6%B1%82%E8%AF%81%EF%BC%9AEB%3DOA)
如图E为菱形ABCD边BC上一点且AB=AE AE交BD于O 且∠DAE=2∠BAE 求证:EB=OA
如图E为菱形ABCD边BC上一点且AB=AE AE交BD于O 且∠DAE=2∠BAE 求证:EB=OA
如图E为菱形ABCD边BC上一点且AB=AE AE交BD于O 且∠DAE=2∠BAE 求证:EB=OA
证明:∵∠DAE=2∠BAE ,AD‖BC∴∠AEB=∠DAE
∵AB=AE∴∠ABE=∠AEB∴∠ABE=∠DAE=2∠BAE
∴设∠BAE=x°所以∠ABE=∠AEB=2x°
∴x+2x+2x=180,x=36°
∴∠ABE=∠AEB=∠DAE=72°
∴∠BAD=108°
∵是菱形∴AB=AD,∠ABD=∠ADB=36°∴∠DOA=72°,△DOA是等腰三角形
∴证明△ABE和△DOA全等就行
∠BAE=∠ADO,AB=DA,∠ABE=∠DAO
∴全等
∴OA=EB
证明:∵∠DAE=2∠BAE ,AD‖BC∴∠AEB=∠DAE
∵AB=AE∴∠ABE=∠AEB∴∠ABE=∠DAE=2∠BAE
∴设∠BAE=x°所以∠ABE=∠AEB=2x°
∴x+2x+2x=180,x=36°
∴∠ABE=∠AEB=∠DAE=72°
∴∠BAD=108°
∵是菱形∴AB=AD,∠ABD=∠ADB=36°∴∠DOA=72°,△D...
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证明:∵∠DAE=2∠BAE ,AD‖BC∴∠AEB=∠DAE
∵AB=AE∴∠ABE=∠AEB∴∠ABE=∠DAE=2∠BAE
∴设∠BAE=x°所以∠ABE=∠AEB=2x°
∴x+2x+2x=180,x=36°
∴∠ABE=∠AEB=∠DAE=72°
∴∠BAD=108°
∵是菱形∴AB=AD,∠ABD=∠ADB=36°∴∠DOA=72°,△DOA是等腰三角形
∴证明△ABE和△DOA全等就行
∠BAE=∠ADO,AB=DA,∠ABE=∠DAO
∴全等
∴OA=EB
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分析:要EB=OA,证它们所在的三角形全等,即△AOD≌△BEA 证明:∵四边形ABCD为菱形,∴AD‖BC,AD=BA, ∠ABC=∠ADC=2∠ADB ∴∠DAE=∠AEB ∵AB=AE,∴∠ABC=∠AEB ∴∠ABC=∠DAE ∵∠DAE=2∠BAE,∴∠BAE=∠ADB 又∵AD=BA ∴△AOD≌△BEA ∴AO=BE