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 F
Something is not right here, ....
If D is the midpoint of BC , then AD is clearly a median
you then say, G is the midpoint of AD such that
AG:GD = 2:1
From the 2:1 ratio we know that G must be the centroid, which of course cannot be the midpoint.
To prove that B, G, and E are collinear, we can use vector algebra. Here's how:
1. Let's start by defining the vectors:
AC = C - A
BC = C - B
2. Since D is the midpoint of BC, we can express D as the average of B and C:
D = (B + C) / 2
3. Similarly, since E is the midpoint of AC, we can express E as the average of A and C:
E = (A + C) / 2
4. Now, we can express G as the midpoint of AD, using the given ratio:
G = A + (2/3)(D - A)
= (3A + 2D) / 3
= (3A + 2(B + C) / 2) / 3
= (3A + B + C) / 3
5. B, G, and E are collinear if the vectors BG and GE are parallel. We can check this by showing that the cross product of BG and GE is zero:
BG x GE = (B - G) x (G - E)
= (B - (3A + B + C) / 3) x ((3A + B + C) / 3 - (A + C) / 2)
= (2B - 3A - C) x (3A + B - C) / 3
6. To simplify further, we can distribute the cross product:
BG x GE = (2B - 3A - C) x (3A + B - C) / 3
= (2B x (3A + B - C) - 3A x (3A + B - C) - C x (3A + B - C)) / 3
7. Expand each term in the cross product:
BG x GE = (6AB + 2B^2 - 6A^2 - 2AB - 3A^2 - 3AB - 3AC - 3BA - BC + 3AC + CB) / 3
= (-6A^2 - 2A^2 - 6AB - 2AB - 3BA + CB - BC) / 3
8. Simplify the expression:
BG x GE = (-8A^2 - 11AB + CB - BC) / 3
9. Now, we know that vectors AB = p and BC = q. We can substitute these values into the expression:
BG x GE = (-8A^2 - 11AB + CB - BC) / 3
= (-8A^2 - 11p + q - q) / 3
= (-8A^2 - 11p) / 3
10. Since the cross product is zero if and only if the numerator is zero, we need to show that:
-8A^2 - 11p = 0
11. Since this equation involves the vector p, and we don't know anything about its relation to the triangle ABC, we cannot prove that B, G, and E are collinear based on the given information.
To prove the same result for point C, G, and F, the approach is similar. However, this time, the points and vectors involved will be different.