如图,已知AB=AE,BC=ED,∠B=∠E,AF⊥CD,F为垂足,试说明CF=DF.AB EC F D..........................................A............................B.........................EC.....F .....D
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![如图,已知AB=AE,BC=ED,∠B=∠E,AF⊥CD,F为垂足,试说明CF=DF.AB EC F D..........................................A............................B.........................EC.....F .....D](/uploads/image/z/2697926-14-6.jpg?t=%E5%A6%82%E5%9B%BE%2C%E5%B7%B2%E7%9F%A5AB%3DAE%2CBC%3DED%2C%E2%88%A0B%3D%E2%88%A0E%2CAF%E2%8A%A5CD%2CF%E4%B8%BA%E5%9E%82%E8%B6%B3%2C%E8%AF%95%E8%AF%B4%E6%98%8ECF%3DDF.AB+EC+F+D..........................................A............................B.........................EC.....F+.....D)
如图,已知AB=AE,BC=ED,∠B=∠E,AF⊥CD,F为垂足,试说明CF=DF.AB EC F D..........................................A............................B.........................EC.....F .....D
如图,已知AB=AE,BC=ED,∠B=∠E,AF⊥CD,F为垂足,试说明CF=DF.
A
B E
C F D
..........................................A..
..........................B.........................E
C.....F .....D
如图,已知AB=AE,BC=ED,∠B=∠E,AF⊥CD,F为垂足,试说明CF=DF.AB EC F D..........................................A............................B.........................EC.....F .....D
做辅助线AC,AD
因为 AB=AE,BC=ED,∠B=∠E(边角边)
所以 三角形ABC = 三角形AED (全等)
所以 AC=AD
所以 三角形 ACD 是等腰三角形
所以 ∠ACF=∠ADF
因为 AF⊥CD
所以 ∠AFC=∠AFD=90°
因为 AC=AD,∠ACF=∠ADF,∠AFC=∠AFD(角角边)
所以 三角形AFC = 三角形AFD (全等)
所以 CF=DF
做辅助线AC,AD,因为AB=AE,BC=ED,∠B=∠E,所以三角形ABC与三角形AED全等(边角边)所以AC=AD。在直角三角形AFC与AFD中,因为AF=AF,AC=AD,所以两个直角三角形全等,所以CF=DF