Using vectors, prove that the lines from each vertex of a triangle to the midpoint of the opposite side cross at a single point. The picture is as shown below:
We solve this by constructing parametric equations for the midpoint lines. The first one starts at O and travels along the vector a + (b-a)/2, which is the diagonal of the parallelogram formed by a and b. The equation is:
u[a + (b-a)/2]where u is the parameter. Similarly, construct the line extending from A to its opposing side. It starts from A and travels along the vector b/2 - a. The equation is:
a + v(b/2 - a)At the point where the vectors cross, these are equal:
u[a + (b-a)/2] = a + v(b/2 - a)I had a little trouble at this point, and I must confess I peeked at the answer. We can solve this by considering that both the a and the b terms must balance. (Even though a isn't perpendicular to b, the part of b which is perpendicular to a is a constant fraction k of the whole).
This leads to:
½ k ub = ½ k vb
u = v = ⅔ua - ½ ua = a - va
½ u + v = 1
u = v = ⅔Given this suggested solution u = v = ⅔ , we can easily verify that
u[a + (b-a)/2]⅔[a + (b-a)/2] = ⅔ a + ⅓ b - ⅓ a = ⅓ (a + b)a + v(b/2 - a)a + ⅔(b/2 - a) = a + ⅓ b -⅔ a = ⅓ (a + b)Now consider the third side. The midpoint line starts from B and travels along the vector a/2 - b. The equation is:
b + w(a/2 - b)We observe that for w = ⅔ this becomes:
b + w(a/2 - b)b + ⅔(a/2 - b)b + ⅓ a - ⅔ b = ⅓ (a + b)Thus, all three lines intersect at the same point.
I haven't got an extension to the general case just yet (i.e. any point, not just the midpoint).
