4. 已知:如图,点E、F分别在AB、CD上,$\frac{AE}{EB}= \frac{DF}{FC}$.求证:(1) $\frac{AB}{EB}= \frac{DC}{FC}$. (2) $\frac{AB}{AE}= \frac{DC}{DF}$.

答案:
证明:
(1)
∵$\frac {AE}{EB}=\frac {DF}{FC},$
∴$\frac {AE}{EB}+1=\frac {DF}{FC}+1,$
∴$\frac {AB}{EB}=\frac {DC}{FC}.$
(2)
∵$\frac {AE}{EB}=\frac {DF}{FC},$
∴$\frac {EB}{AE}=\frac {FC}{DF},$
∴$\frac {EB}{AE}+1=\frac {FC}{DF}+1,$
∴$\frac {EB+AE}{AE}=\frac {FC+DF}{DF},$
∴$\frac {AB}{AE}=\frac {DC}{DF}.$
(1)
∵$\frac {AE}{EB}=\frac {DF}{FC},$
∴$\frac {AE}{EB}+1=\frac {DF}{FC}+1,$
∴$\frac {AB}{EB}=\frac {DC}{FC}.$
(2)
∵$\frac {AE}{EB}=\frac {DF}{FC},$
∴$\frac {EB}{AE}=\frac {FC}{DF},$
∴$\frac {EB}{AE}+1=\frac {FC}{DF}+1,$
∴$\frac {EB+AE}{AE}=\frac {FC+DF}{DF},$
∴$\frac {AB}{AE}=\frac {DC}{DF}.$
5. 已知△ABC和△A'B'C'中,$\frac{AB}{A'B'}= \frac{BC}{B'C'}= \frac{CA}{C'A'}= \frac{2}{3}$,且A'B'+B'C'+C'A'= 24厘米,求△ABC的周长.
答案:
解:
∵$\frac {AB}{A′B′}=\frac {BC}{B′C′}=\frac {CA}{C′A′}=\frac {2}{3},$
∴△ABC∽△A′B′C′,
∴△ABC的周长:△A′B′C′的周长=2:3.
∵$A′B′+B′C′+C′A′=24\ \mathrm {cm},$
∴△ABC的周长$=16\ \mathrm {cm}.$
∵$\frac {AB}{A′B′}=\frac {BC}{B′C′}=\frac {CA}{C′A′}=\frac {2}{3},$
∴△ABC∽△A′B′C′,
∴△ABC的周长:△A′B′C′的周长=2:3.
∵$A′B′+B′C′+C′A′=24\ \mathrm {cm},$
∴△ABC的周长$=16\ \mathrm {cm}.$
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