2025年通城学典课时作业本八年级数学上册苏科版苏州专版第13页答案
8. 如图,在△ABC和△EDF中,$ ∠B = ∠D = 90° $,点A,E,C,F在同一条直线上,$ AE = CF $,BC的延长线交DF于点M,$ ∠MCF = ∠F $.求证:$ BC = DF $.
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答案

8.
∵∠MCF = ∠F,∠MCF = ∠ACB,
∴∠ACB = ∠F.
∵AE = CF,
∴AE + EC = CF + EC,即AC = EF.在△ABC和△EDF中,$\begin{cases}∠B = ∠D,\\∠ACB = ∠F,\\AC = EF,\end{cases}(AAS),$
∴BC = DF
9. (2023·营口改编)如图,点A,B,C,D在同一条直线上,点E,F分别在直线AB的两侧,且$ AE = BF $,$ AE// BF $,$ ∠ECD = ∠FDC $,连接CF,DE.
(1)求证:$ CE = DF $;
(2)若$ AB = 8 $,$ AC = 2 $,求CD的长.
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答案

9.
(1)
∵AE//BF,
∴∠A = ∠B.
∵∠ACE + ∠ECD = 180°,∠BDF + ∠FDC = 180°,∠ECD = ∠FDC,
∴∠ACE = ∠BDF.在△ACE和△BDF中,$\begin{cases}∠A = ∠B,\\∠ACE = ∠BDF,\\AE = BF,\end{cases}$
∴△ACE≌△BDF(AAS),
∴CE = DF
(2)由
(1)知,△ACE≌△BDF,
∴AC = BD = 2.
∵AB = 8,
∴CD = AB - AC - BD = 4,
∴CD的长为4
10. 如图,点A,F,E,C在同一条直线上,$ AB// CD $,$ ∠ABE = ∠CDF $,$ AF + AE = AC $.
(1)① △ABE≌
△CDF

② △BCE≌
△DAF

③ △ABC≌
△CDA
.
(2)对(1)中的①②加以证明.
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答案

10.
(1)①△CDF ②△DAF ③△CDA
(2)①
∵AB//CD,
∴∠BAE = ∠DCF.
∵AF + AE = AC,
∴AE = AC - AF = CF.在△ABE和△CDF中,$\begin{cases}∠ABE = ∠CDF,\\∠BAE = ∠DCF,\\AE = CF,\end{cases}$
∴△ABE≌△CDF(AAS) ②
∵△ABE≌△CDF,
∴∠AEB = ∠CFD,BE = DF,
∴∠BEC = ∠DFA.
∵AF + AE = AC,CE + AE = AC,
∴AF = CE.在△BCE和△DAF中,$\begin{cases}BE = DF,\\∠BEC = ∠DFA,\\CE = AF,\end{cases}$
∴△BCE≌△DAF(SAS)