The points D, E and F divide the sides BC, CA and AB of a triangle in the ratio 1:4, 3:2 and 3:7 respectively. Show that there exists a point G such that the sum of the vectors AD, BE, CF is parallel to CG. What is the ratio in which G divides AB?
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Expert's answer
2021-01-28T19:04:12-0500
In triangle ABC, let A is zero vector.
B is b and C is c . D divides BC in 1:4 thus D is 5c+4b . Similarly E is 52c. F is 103b .
Now adding AD +BE+CF, we get
5c+4b−0+52c−b+103b−c
=10b−52c
Now we know that point G divides the AB so that let it's divides in ratio 1:k then G=k+1b and
CG=k+1b−c .
Now for k=3, CG=4b−c.
(2/5)(CG)=10b−52c .
We can see that addition of AD +BE +CF is equals to 52CG. Hence it proves that the above sum is parallel to CG and also G divides the AB in 1:3 .
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