如图,在△ABC中,∠BAD=∠DAC,BE⊥AC于E,交AD于F,式说明:∠AFE=1/2(∠ABC+∠C) 但不要太多.
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如图,在△ABC中,∠BAD=∠DAC,BE⊥AC于E,交AD于F,式说明:∠AFE=1/2(∠ABC+∠C) 但不要太多.
如图,在△ABC中,∠BAD=∠DAC,BE⊥AC于E,交AD于F,式说明:∠AFE=1/2(∠ABC+∠C) 但不要太多.
如图,在△ABC中,∠BAD=∠DAC,BE⊥AC于E,交AD于F,式说明:∠AFE=1/2(∠ABC+∠C) 但不要太多.
∠ABC+∠C=180-∠BAC
∠ABC+∠C=180-2∠DAC=2(90-∠DAC)
因为BE⊥AC所以90-∠DAC=∠AFE
所以∠ABC+∠C=2∠AFE
所以:∠AFE=1/2(∠ABC+∠C)
∵BE⊥AC
∴∠FEA=90
∴∠AFE=180-90-∠DAC=90-∠DAC
∵∠DAC=∠BAD=1/2∠A
∠A=180-(∠ABC+∠C)
∴∠AFE=90-1/2∠A
=90-【180-(∠ABC+∠C)】1/2
=90-90+1/2∠ABC+1/2∠C
=1/2(∠ABC+∠C)
∠ABC+∠C+∠BAC=180
所以∠ABC+∠ACB=180-∠BAC=180-2∠CAD
BE⊥AC于E
∠CAD+∠AFE=90 两边乘以2
2∠CAD+2∠AFE=180
2个等式代换下就可以得到要证明的AFE=1/2(∠ABC+∠C)
∠AFE=1/2(∠ABC+∠C)也就是∠AFE=1/2(180°-∠BAC)也就是90°-1/2∠BAC最后就是∠AFE=90°-∠EAF,因为BC⊥AC,∠BEA=90°所以∠AFE=90°-∠EAF,再换回去,就是:∠AFE=1/2(∠ABC+∠C)。。。
呵呵、、