Answer to Question #158853 in Geometry for Angie

Question

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?

Answer 1

In triangle ABC, let A is zero vector.

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

Now adding AD +BE+CF, we get

$\frac{\overrightarrow{c}+\overrightarrow{4b}}{5} - \overrightarrow{0} + \frac{\overrightarrow{2c}}{5} - \overrightarrow{b} + \frac{\overrightarrow{3b}}{10} - \overrightarrow{c}$

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

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

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

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

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

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