如图AB=AE,AC=AD,∠BAD=∠CAE=90°.AH⊥BC于点H,HA的延长线交DE于G.求证:GD=GE.
来源:学生作业帮助网 编辑:作业帮 时间:2024/06/30 00:57:46
![如图AB=AE,AC=AD,∠BAD=∠CAE=90°.AH⊥BC于点H,HA的延长线交DE于G.求证:GD=GE.](/uploads/image/z/12514536-0-6.jpg?t=%E5%A6%82%E5%9B%BEAB%3DAE%2CAC%3DAD%2C%E2%88%A0BAD%3D%E2%88%A0CAE%3D90%C2%B0.AH%E2%8A%A5BC%E4%BA%8E%E7%82%B9H%2CHA%E7%9A%84%E5%BB%B6%E9%95%BF%E7%BA%BF%E4%BA%A4DE%E4%BA%8EG.%E6%B1%82%E8%AF%81%3AGD%3DGE.)
如图AB=AE,AC=AD,∠BAD=∠CAE=90°.AH⊥BC于点H,HA的延长线交DE于G.求证:GD=GE.
如图AB=AE,AC=AD,∠BAD=∠CAE=90°.AH⊥BC于点H,HA的延长线交DE于G.求证:GD=GE.
如图AB=AE,AC=AD,∠BAD=∠CAE=90°.AH⊥BC于点H,HA的延长线交DE于G.求证:GD=GE.
在BC上截取BG'=AG
∵∠BAD=∠CAE=∠AHB=∠AHC=90°
∴∠BAH+∠ABC=∠BAH+∠DAG=∠CAH+∠BCA=∠CAH+∠EAG=90°
∴∠CBA=∠DAG,∠BCA=∠EAG
又∵AB=AD,AG=BG'
∴△ABG'≌△ADG(SAS)
∴DG=AG',∠DGA=∠BG'A
∴∠EGA=∠CG'A
又∵∠BCA=∠EAG,AC=AE
∴△ACG'≌△AEG(AAS)
∴GE=AG'=GD
如图,AB/AD=BC/DE=AC/AE,求证:∠BAD=∠CAE
如图AB=AD ∠BAD=∠CAE AC=AE 求证AB=AD
如图,∠B=∠E,∠BAD=∠EAC,AC=AD,求证:AB=AE
已知:如图AB=AD,AC=AE,∠BAD=∠CAE ,求证:OA平分∠BOE
如图已知ab=ad,ac=ae∠bad=∠eac是说明bc=de
1、如图1111,AB=AD,∠BAD=∠CAE,AC=AE,求证:DE=BC
如图,AB=AD,AC=AE,∠BAD=∠CAE,连接BC、DE,求证:bc=de
如图,已知AB=AC,AD=AE,∠BAD=∠CAE.求证:BE=CD.
已知:如图,AB=AD,AC=AE,∠BAD=∠CAE,求证:BC=DE
如图,已知AB=AC,AD=AE,∠BAD=∠CAE,求证:BE=CD
如图,AB=AC,∠BAD=30°,AD=AE,则∠EDC=________.
已知:如图,AB/AD=BC/DE=AC/AE,求证:∠BAD=∠CAE
如图,已知AB分之AE = BC分之ED = AC分之AD 证明∠BAD=∠CAE
如图,AB平行AC,∠BAD=∠CAE,AD=AE,求证:△ABE≌△ACD.
已知:如图,AD平行BC,AE平分∠BAD,AE⊥BE;说明:AD+BC=AB
已知:如图,AD平行于BC,AE平分∠BAD,AE⊥BE;说明:AD+BC=AB
已知:如图,AD∥BC,AE平分∠BAD,AE⊥BE;说明:AD+BC=AB.
已知:如图,AD平行于BC,AE平分∠BAD,AE⊥BE;说明:AD+BC=AB