Question

The points D, E and F divide the sides BC, CA and AB of a triangle in the ratios

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 ratio does G divide AB?​

Answer 1

Given :

The points D, E and F divide the sides BC, CA and AB of a triangle in the ratios  1 : 4, 3 : 2 and 3 : 7 respectively.

To find :

Show that there exists a point G such that the sum  of the vectors AD, BE, CF is parallel to CG.

What ratio does G divide AB?

Solution :

By section formula, we got points D, E, and F

AD = (c + 4b)/5 - 0

AD = (c+4b)/5

BE = 2c/5 - b

CF = 3b/10 - c

So, AD + BE + CF = (c+4b)/5 + (2c/5 - b) + (3b/10 - c)

AD + BE + CF = b/10 - 2c/5

We know, EA = -2c/5

Let us call a point K on AB such that AK = b/10

Then, we have AK + EA = AD + BE + CF

AK + EA = EK

Now, let a point G on AB such that CG is parallel to EK

Now, since CG is parallel to EK

ΔAEK is similar to ΔACG

So, AC/AE = AG/AK

c/(2c/5) = AG/(b/10)

So, AG = (5/2) × (b/10)

AG = b/4

So, point G = b/4

Since, AG < AB

So, there exist a point G such that AD + BE + CF is parallel to CG

Now, let G divide AB in ratio 1:k

By section formula,

G = b/4 = (1×b + k×0)/(1+k)

b/(k+1) = b/4

k = 3

So, G divide AB in ratio 1:3