已知:如图所示,∠ABC=∠ADC,BF和DE分别平分∠ABC和∠ADC,∠AED=∠EDC.求证:ED∥BF.
证明:∵BF和DE分别平分∠ABC和∠ADC(已知)
∴∠EDC={#blank#}1{#/blank#} ∠ADC,
∠FBA={#blank#}2{#/blank#} ∠ABC(角平分线定义).
又∵∠ADC=∠ABC(已知),
∴∠{#blank#}3{#/blank#} =∠FBA(等量代换).
又∵∠AED=∠EDC(已知),
∴∠{#blank#}4{#/blank#}=∠{#blank#}5{#/blank#} (等量代换),
∴ED∥BF{#blank#}6{#/blank#} .
![](http://tikupic.21cnjy.com/c9/45/c9452cd33cc3a736ba3ecb43d396e589.png)