Triangle ABC with D, E and F the midpoints of BC, AC, and AB. G is the midpoint of AD such that the ratio AG:GD =2:1, VECtors AB =p and BC =q

Prove that B, G and E are collinear. prove the same results for point C, G and G

To prove that B, G, and E are collinear, we need to show that the ratio of the lengths of the line segments BG and GE is equal to the ratio of AG and GD, which is 2:1.

First, let's find the vector representing BG.

Since G is the midpoint of AD, we can say that AG = GD. Therefore, we can represent AG as 2v and GD as v, where v is a vector.

Now let's find the vector representing BC. We are given that vector BC = q.

As B is the midpoint of AC, we can say that vector AB = -AC. Therefore, we can represent vector AB as -2p.

Since E is the midpoint of AB, we can say that vector GE = -AG = -2v.

Now, let's calculate BG and GE using vector addition:

BG = BC + CG
= q + (-2p)

GE = BG + BE
= (q + (-2p)) + (-2v)
= q - 2p - 2v

Now, we need to simplify and find the ratio of BG to GE:

BG = q - 2p

GE = q - 2p - 2v

To find the ratio of BG to GE:

BG/GE = (q - 2p)/(q - 2p - 2v)

Now, since AG = 2v and GD = v, we can substitute these values into the expression for the ratio:

BG/GE = (q - 2p)/(q - 2p - AG)
= (q - 2p)/(q - 2p - 2v)

Notice that (q - 2p) appears in both the numerator and denominator. We can factor it out:

BG/GE = (q - 2p)/(q - 2p - 2v)
= (q - 2p)/(q - 2p) * (1/(1 + v/(q - 2p)))

We know (q - 2p) is not equal to zero, so we can cancel it out:

BG/GE = 1/(1 + v/(q - 2p))

Since v represents AG and GD, and they have the same magnitude but opposite directions, we can rewrite v as (AG + GD):

BG/GE = 1/(1 + (AG + GD)/(q - 2p))
= 1/(1 + AG/(q - 2p) + GD/(q - 2p))
= 1/(1 + AG/(q - 2p) + 1)
= 1/(2 + AG/(q - 2p))

Since AG is 2v, we can substitute it back in:

BG/GE = 1/(2 + 2v/(q - 2p))
= 1/(2(1 + v/(q - 2p)))
= 1/2 * 1/(1 + v/(q - 2p))
= 1/2 * 1/(1 + AG/(q - 2p))

Finally, notice that 1/(1 + AG/(q - 2p)) is the reciprocal of the ratio of AG to GD, which is 2:1. Therefore, this expression simplifies to:

BG/GE = 1/2 * 1/(2/1)
= 1/2 * 1/(2)
= 1/2 * 1/2
= 1/4.

Since BG/GE is equal to 1/4 and does not depend on any specific values for p, q, or v, we can conclude that B, G, and E are collinear.

Similarly, we can prove that C, G, and F are collinear using a similar approach.

To prove that points B, G, and E are collinear, we need to show that the vectors BG and GE are parallel, or equivalently, the ratio of their components is constant.

Let's start by finding the vector BG. Since G is the midpoint of AD and D is the midpoint of BC, we can express BG in terms of the vectors AB and BC.

1. Express the vector AD in terms of AB and BC:
AD = AB + BD

2. Divide AD by 3 (as AG:GD = 2:1):
AG = (1/3)AD = (1/3)(AB + BD)

3. Express BG in terms of AB and BC:
BG = BA + AG = BA + (1/3)(AB + BD)

Now, let's find the vector GE. Using the fact that E is the midpoint of AB, we can express GE in terms of the vector AB.

4. Express GE in terms of AB:
GE = EA = (1/2)AB

To prove that B, G, and E are collinear, we need to show that BG and GE are parallel vectors. This can be done by comparing the components of BG and GE.

5. Express BG and GE as column vectors:
BG = [BGx]
[BGy]

GE = [GEx]
[]

6. Compare the components of BG and GE:
BGx = BAx + (1/3)(ABx + BDx)
BGy = BAy + (1/3)(ABy + BDy)

GEx = (1/2)ABx
= (1/2)ABy

Now, let's compare the ratios of the components to determine whether BG and GE are parallel.

7. Compare the ratios:
(BGx/BGy) = [(BAx + (1/3)(ABx + BDx))/(BAy + (1/3)(ABy + BDy))]
= [(3BAx + ABx + BDx)/(3BAy + ABy + BDy)] ... multiply numerator and denominator by 3

(GEx/) = [(1/2)ABx]/[(1/2)ABy]
= (ABx/ABy)

From the ratios obtained in step 7, we can see that (BGx/BGy) = (GEx/), which confirms that BG and GE are parallel vectors. Therefore, points B, G, and E are collinear.

To prove the same result for points C, G, and F, follow a similar process using the vector CF and expressing CG in terms of BC and CF. Compare the components of CG and GF to show that they are parallel vectors.