Question

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Answer 1

Bài 3:

\(\dfrac{CN}{AN}=\dfrac{1}{3}\)

=>AN=3CN

Ta có: CN+NA=CA

=>CN+3CN=CA

=>CA=4CN

=>\(\dfrac{CN}{CA}=\dfrac{1}{4}\)

CB=4CM

=>\(\dfrac{CM}{CB}=\dfrac{1}{4}\)

Xét ΔCAB có \(\dfrac{CM}{CB}=\dfrac{CN}{CA}\left(=\dfrac{1}{4}\right)\)

nên MN//AB

Bài 4:

a: Xét ΔOBA có ED//BA

nên \(\dfrac{OE}{OB}=\dfrac{OD}{OA}\left(1\right)\)

b: 

Xét ΔOAC có DF//AC

nên \(\dfrac{OF}{OC}=\dfrac{OD}{OA}\left(2\right)\)

c:

Từ (1),(2) suy ra \(\dfrac{OE}{OB}=\dfrac{OF}{OC}\)

Xét ΔOBC có \(\dfrac{OE}{OB}=\dfrac{OF}{OC}\)

nên EF//BC