4. (1)如图,已知:$AB// CD$,$BE// CF$. 求证:$∠1=∠2$. 把以下证明过程补充完整.

证明 $\because AB// CD$,
$\therefore ∠ABC=∠BCD$(
$\because BE// CF$,
$\therefore$
$\therefore$
$\therefore ∠1=∠2$.
(2)如图,已知:$AB// CD$,$∠1=∠2$. 求证:$BE// CF$.
(3)如图,已知:$∠ABC=∠BCD$,$BE$、$CF$分别平分$∠ABC$和$∠BCD$. 求证:$BE// CF$.
证明 $\because AB// CD$,
$\therefore ∠ABC=∠BCD$(
两直线平行,内错角相等
).$\because BE// CF$,
$\therefore$
∠EBC
=∠BCF
(两直线平行,内错角相等
).$\therefore$
∠ABC
−∠EBC
=∠BCD
−∠BCF
.$\therefore ∠1=∠2$.
(2)如图,已知:$AB// CD$,$∠1=∠2$. 求证:$BE// CF$.
(3)如图,已知:$∠ABC=∠BCD$,$BE$、$CF$分别平分$∠ABC$和$∠BCD$. 求证:$BE// CF$.
答案
4. (1) 两直线平行,内错角相等;∠EBC;∠BCF;两直线平行,内错角相等;∠ABC;∠EBC;∠BCD;∠BCF.
(2)
∵ AB//CD,
∴ ∠ABC = ∠BCD(两直线平行,内错角相等).
∵ ∠1 = ∠2,
∴ ∠ABC - ∠1 = ∠BCD - ∠2.
∴ ∠EBC = ∠BCF.
∴ BE//CF(内错角相等,两直线平行).
(3)
∵ BE 平分∠ABC,
∴ ∠CBE = $\frac{1}{2}$∠ABC.
同理,∠BCF = $\frac{1}{2}$∠BCD.
∵ ∠ABC = ∠BCD,
∴ ∠CBE = ∠BCF.
∴ BE//CF(内错角相等,两直线平行).
(2)
∵ AB//CD,
∴ ∠ABC = ∠BCD(两直线平行,内错角相等).
∵ ∠1 = ∠2,
∴ ∠ABC - ∠1 = ∠BCD - ∠2.
∴ ∠EBC = ∠BCF.
∴ BE//CF(内错角相等,两直线平行).
(3)
∵ BE 平分∠ABC,
∴ ∠CBE = $\frac{1}{2}$∠ABC.
同理,∠BCF = $\frac{1}{2}$∠BCD.
∵ ∠ABC = ∠BCD,
∴ ∠CBE = ∠BCF.
∴ BE//CF(内错角相等,两直线平行).
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