如图在RT三角形ABC中角BAC=90°AB=AC,AE是过A点的一条直线且B点和C点在AE的两侧BD⊥AE于点D AE垂直如图1,在RT三角形ABC中,角BAC=90°,AB=AC,AE是过A点的一条直线,且B点和C点在AE的两侧,BD⊥AE于点D,CE⊥AE于
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![如图在RT三角形ABC中角BAC=90°AB=AC,AE是过A点的一条直线且B点和C点在AE的两侧BD⊥AE于点D AE垂直如图1,在RT三角形ABC中,角BAC=90°,AB=AC,AE是过A点的一条直线,且B点和C点在AE的两侧,BD⊥AE于点D,CE⊥AE于](/uploads/image/z/916326-54-6.jpg?t=%E5%A6%82%E5%9B%BE%E5%9C%A8RT%E4%B8%89%E8%A7%92%E5%BD%A2ABC%E4%B8%AD%E8%A7%92BAC%3D90%C2%B0AB%3DAC%2CAE%E6%98%AF%E8%BF%87A%E7%82%B9%E7%9A%84%E4%B8%80%E6%9D%A1%E7%9B%B4%E7%BA%BF%E4%B8%94B%E7%82%B9%E5%92%8CC%E7%82%B9%E5%9C%A8AE%E7%9A%84%E4%B8%A4%E4%BE%A7BD%E2%8A%A5AE%E4%BA%8E%E7%82%B9D+AE%E5%9E%82%E7%9B%B4%E5%A6%82%E5%9B%BE1%2C%E5%9C%A8RT%E4%B8%89%E8%A7%92%E5%BD%A2ABC%E4%B8%AD%2C%E8%A7%92BAC%3D90%C2%B0%2CAB%3DAC%2CAE%E6%98%AF%E8%BF%87A%E7%82%B9%E7%9A%84%E4%B8%80%E6%9D%A1%E7%9B%B4%E7%BA%BF%2C%E4%B8%94B%E7%82%B9%E5%92%8CC%E7%82%B9%E5%9C%A8AE%E7%9A%84%E4%B8%A4%E4%BE%A7%2CBD%E2%8A%A5AE%E4%BA%8E%E7%82%B9D%2CCE%E2%8A%A5AE%E4%BA%8E)
如图在RT三角形ABC中角BAC=90°AB=AC,AE是过A点的一条直线且B点和C点在AE的两侧BD⊥AE于点D AE垂直如图1,在RT三角形ABC中,角BAC=90°,AB=AC,AE是过A点的一条直线,且B点和C点在AE的两侧,BD⊥AE于点D,CE⊥AE于
如图在RT三角形ABC中角BAC=90°AB=AC,AE是过A点的一条直线且B点和C点在AE的两侧BD⊥AE于点D AE垂直
如图1,在RT三角形ABC中,角BAC=90°,AB=AC,AE是过A点的一条直线,且B点和C点在AE的两侧,BD⊥AE于点D,CE⊥AE于点E,DE=4则BD-CE=
如图在RT三角形ABC中角BAC=90°AB=AC,AE是过A点的一条直线且B点和C点在AE的两侧BD⊥AE于点D AE垂直如图1,在RT三角形ABC中,角BAC=90°,AB=AC,AE是过A点的一条直线,且B点和C点在AE的两侧,BD⊥AE于点D,CE⊥AE于
∵∠BAC=90°
∴∠BAE+∠CAE=90°
∵BD⊥AE,CE⊥AE
∴∠ADB=∠AEC=90
∴∠BAE+∠ABD=90
∴∠ABD=∠CAE
∵AB=AC
∴△ABD≌△ACE (AAS)
∴AE=BD,AD=CE
∵AE=DE+AD
∴AE=DE+CE
∴BD=DE+CE
∴BD-CE=DE=4
BD-CE=4
证明如下;∵BD⊥AE,CE⊥AE
∴角BDA=∠AEC
∵∠BAC=90°
∴∠BAD+∠CAE=90°
又∵∠BAD+∠ABD=180°-∠BDA=90°
∴∠ABD=∠EAC
∴在△ABD与△CAE中
{∠ADB=∠CEA
∠BAD=∠CAE
AB=AC
∴AD=CE
BD=AE
∴BD-CE=AE-AD=DE=4