如图,∠BAC=90°,AD⊥BC,垂足为D,BE平分∠ABC, AF平分∠DAC,求证:GF∥AC

来源:学生作业帮助网 编辑:作业帮 时间:2024/04/19 05:59:58
如图,∠BAC=90°,AD⊥BC,垂足为D,BE平分∠ABC, AF平分∠DAC,求证:GF∥AC

如图,∠BAC=90°,AD⊥BC,垂足为D,BE平分∠ABC, AF平分∠DAC,求证:GF∥AC
如图,∠BAC=90°,AD⊥BC,垂足为D,BE平分∠ABC, AF平分∠DAC,求证:GF∥AC

如图,∠BAC=90°,AD⊥BC,垂足为D,BE平分∠ABC, AF平分∠DAC,求证:GF∥AC
证明:连接EF.
∵∠BAC=90°,AD⊥BC.
∴∠C+∠ABC=90°,∠C+∠DAC=90°,∠ABC+∠BAD=90°.
∴∠ABC=∠DAC,∠BAD=∠C.
∵BE、AF分别是∠ABC、∠DAC的平分线.
∴∠ABG=∠EBD.
∵∠AGE=∠GAB+∠GBA,∠AEG=∠C+∠EBD,
∴∠AGE=∠AEG,
∴AG=AE,
∵AF是∠DAC的平分线,
∴AO⊥BE,GO=EO,
∵∠ABO=∠FBO
BO=BO
∠AOB=∠FOB=90°
∴△ABO≌△FBO,
∴AO=FO,
∴四边形AGFE是平行四边形,
∴GF∥AE,
即GF∥AC.