摘要:15.证明: ∵AB∥DE.∴∠B=∠DEF.··································································· 1分 ∵BE=CF. ∴BE+CE=CF+CE.即BC=EF.····································· 2分 ∵∠A=∠D.∴△ABC≌△DEF.···························································· 4分 ∴∠ACB=∠F.························································································· 5分
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将以下推理过程及理由补充完整:
证明:∵AB∥DE
∴∠1=
∠AED
(两直线平行,内错角相等
)又∵∠1=∠2
∴∠2=
∠AED
(等量代换
)∴AE∥CD(
内错角相等,两直线平行
).如图,点D,E,F分别是三角形ABC的边BC,CA,AB上的点,AB∥DE,∠1=∠A.求证:FD∥AC.
证明:∵AB∥DE(已知),
∴∠1=
∠BFD
∠BFD
.(两直线平行,内错角相等
两直线平行,内错角相等
)又∠1=∠A(已知),
∴∠A=
∠BFD
∠BFD
.(等量代换
等量代换
)∴FD∥AC.(
同位角相等,两直线平行
同位角相等,两直线平行
)