如图:已知AE=DE,AE垂直DE,AB垂直BC,DC垂直BC.求证AB+CD=BC

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如图:已知AE=DE,AE垂直DE,AB垂直BC,DC垂直BC.求证AB+CD=BC如图:已知AE=DE,AE垂直DE,AB垂直BC,DC垂直BC.求证AB+CD=BC如图:已知AE=DE,AE垂直D

如图:已知AE=DE,AE垂直DE,AB垂直BC,DC垂直BC.求证AB+CD=BC
如图:已知AE=DE,AE垂直DE,AB垂直BC,DC垂直BC.求证AB+CD=BC

如图:已知AE=DE,AE垂直DE,AB垂直BC,DC垂直BC.求证AB+CD=BC
AE⊥DE ,∠AED = 90 = ∠AEB + ∠CED ;∵DC⊥BC ,∴∠CED + ∠EDC = 90 = ∠AEB + ∠CED ,∴∠AEB = ∠EDC ,同理可证∠EAB = ∠DEC ,又∵AE = ED ,∴在△AEB和△EDC中 ,根据“角边角”可证两三角形全等 ,∴对应边相等 ,∴AB = EC ,BE = CD ,∴AB + CD = EC + BE = BC ,即AB + CD = BC ,得证.

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