Question #158853

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?


1
Expert's answer
2021-01-28T19:04:12-0500

In triangle ABC, let A is zero vector.

B is b\overrightarrow{b} and C is c\overrightarrow{c} . D divides BC in 1:4 thus D is c+4b5\frac{\overrightarrow{c} + 4\overrightarrow{b}}{5} . Similarly E is 2c5.\frac{2\overrightarrow{c}}{5}. F is 3b10\frac{3\overrightarrow{b}}{10} .


Now adding AD +BE+CF, we get


c+4b50+2c5b+3b10c\frac{\overrightarrow{c}+\overrightarrow{4b}}{5} - \overrightarrow{0} + \frac{\overrightarrow{2c}}{5} - \overrightarrow{b} + \frac{\overrightarrow{3b}}{10} - \overrightarrow{c}


=b102c5=\frac{\overrightarrow{b}}{10} - \frac{\overrightarrow{2c}}{5}



Now we know that point G divides the AB so that let it's divides in ratio 1:k then G=bk+1\overrightarrow{G} = \frac{\overrightarrow{b}}{k+1} and


CG=bk+1c\overrightarrow{CG} = \frac{\overrightarrow{b}}{k+1} - \overrightarrow{c} .


Now for k=3, CG=b4c.\overrightarrow{CG} = \frac{\overrightarrow{b}}{4} - \overrightarrow{c} .


(2/5)(CG)=b102c5(\overrightarrow{CG}) = \frac{\overrightarrow{b}}{10} - \frac{\overrightarrow{2c}}{5} .


We can see that addition of AD +BE +CF is equals to 2CG5.\frac{2\overrightarrow{CG}}{5}. Hence it proves that the above sum is parallel to CG and also G divides the AB in 1:3 .


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