1. 如图,l₁//l₂//l₃,那么$\frac{AB}{BD}=$
$\frac{EF}{FG}$
$ ,$$\frac{EG}{FG}=$$\frac{AD}{BD}$
$ .$答案
$ \frac {EF}{FG}$
$ \frac {AD}{BD}$
$ \frac {AD}{BD}$
2. 如图,在△ABC中,DE//BC.
(1)若AD = 3,AB = 9,DE = 4,则BC =
(2) 若DE : BC = 2 : 5,则AD : DB = 2 : 3;
(3) 若BC = 7,DE = 4,AE = 8,则EC =
(1)若AD = 3,AB = 9,DE = 4,则BC =
12
;(2) 若DE : BC = 2 : 5,则AD : DB = 2 : 3;
(3) 若BC = 7,DE = 4,AE = 8,则EC =
6
.答案
12
2:3
6
2:3
6
3. 已知:如图,EG//BC,GF//CD.
求证:$\frac{AE}{AB}=\frac{AF}{AD}.$
求证:$\frac{AE}{AB}=\frac{AF}{AD}.$
答案
6
证明:∵EG//BC
∴$\frac {AE}{AB}=\frac {AG}{AC}$
∵GF//CD
∴$\frac {AF}{AD}=\frac {AG}{AC}$
∴$\frac {AE}{AB}=\frac {AF}{AD}$
证明:∵EG//BC
∴$\frac {AE}{AB}=\frac {AG}{AC}$
∵GF//CD
∴$\frac {AF}{AD}=\frac {AG}{AC}$
∴$\frac {AE}{AB}=\frac {AF}{AD}$
1. 如图,在矩形ABCD中,若AB = 3,AC = 5,$\frac{AF}{FC}=\frac{1}{4},$则AE的长为
1
.答案
证明:∵EG//BC
∴$\frac {AE}{AB}=\frac {AG}{AC}$
∵GF//CD
∴$\frac {AF}{AD}=\frac {AG}{AC}$
∴$\frac {AE}{AB}=\frac {AF}{AD}$
1
∴$\frac {AE}{AB}=\frac {AG}{AC}$
∵GF//CD
∴$\frac {AF}{AD}=\frac {AG}{AC}$
∴$\frac {AE}{AB}=\frac {AF}{AD}$
1
2. 如图,在△ABC中,D是边BC的中点,E是边AC上的任意一点,BE交AD于点O.
(1) 当$\frac{AE}{AC}=\frac{1}{2}$时,求$\frac{AO}{AD}$的值;
(2) 当$\frac{AE}{AC}=\frac{1}{3}$时,求$\frac{AO}{AD}$的值;
(3) 当$\frac{AE}{AC}=\frac{1}{4}$时,求$\frac{AO}{AD}$的值;
(4) 当$\frac{AE}{AC}=\frac{1}{n + 1}$时,试猜想$\frac{AO}{AD}$的值,并证明你的猜想.
(1) 当$\frac{AE}{AC}=\frac{1}{2}$时,求$\frac{AO}{AD}$的值;
(2) 当$\frac{AE}{AC}=\frac{1}{3}$时,求$\frac{AO}{AD}$的值;
(3) 当$\frac{AE}{AC}=\frac{1}{4}$时,求$\frac{AO}{AD}$的值;
(4) 当$\frac{AE}{AC}=\frac{1}{n + 1}$时,试猜想$\frac{AO}{AD}$的值,并证明你的猜想.
答案
解:过点D作DF//BE,交AC于点F
∵DF//BE
∴$\frac {CF}{CE}=\frac {CD}{BC}=\frac {1}{2}$
∴点F 为CE的中点
∵DF//BE
∴$\frac {AO}{AD}=\frac {AE}{AF}$
(1)∵$\frac {AE}{AC}=\frac {1}{2}$
∴$\frac {AE}{AF}=\frac {2}{3}$
∴$\frac {AO}{AD}=\frac {2}{3}$
(2)∵$\frac {AE}{AC}=\frac {1}{3}$
∴$\frac {AE}{AF}=\frac {1}{2}$
∴$\frac {AO}{AD}=\frac {1}{2}$
(3)∵$\frac {AE}{AC}=\frac {1}{4}$
∴$\frac {AE}{AF}=\frac {2}{5}$
∴$\frac {AO}{AD}=\frac {2}{5}$
(4)∵$\frac {AE}{AC}=\frac {1}{n+1}$
∴$\frac {AE}{AF}=\frac {2}{n+2}$
∴$\frac {AO}{AD}=\frac {2}{n+2}$
∵DF//BE
∴$\frac {CF}{CE}=\frac {CD}{BC}=\frac {1}{2}$
∴点F 为CE的中点
∵DF//BE
∴$\frac {AO}{AD}=\frac {AE}{AF}$
(1)∵$\frac {AE}{AC}=\frac {1}{2}$
∴$\frac {AE}{AF}=\frac {2}{3}$
∴$\frac {AO}{AD}=\frac {2}{3}$
(2)∵$\frac {AE}{AC}=\frac {1}{3}$
∴$\frac {AE}{AF}=\frac {1}{2}$
∴$\frac {AO}{AD}=\frac {1}{2}$
(3)∵$\frac {AE}{AC}=\frac {1}{4}$
∴$\frac {AE}{AF}=\frac {2}{5}$
∴$\frac {AO}{AD}=\frac {2}{5}$
(4)∵$\frac {AE}{AC}=\frac {1}{n+1}$
∴$\frac {AE}{AF}=\frac {2}{n+2}$
∴$\frac {AO}{AD}=\frac {2}{n+2}$
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