Example 6-1. Show that the medians of a triangle meet at a trisection point.
Figure 6-4. Constructing medians of a triangle
Solution: Let the sides of a triangle with vertices α, β, γ be denoted by the vectors
A, B, and A−B as illustrated in the figure 6-4. Further, let α, β, γ denote the
vectors from the respective vertices of α, β, γ to the midpoints of the opposite sides.
By using the above definitions one can construct the following vector equations
A+α=^1
2
B B+^1
2
(A−B) = β B+γ =^1
2
A. (6 .5)
Let the vectors αand βintersect at a point designated by P, Similarly, let the vectors
βand γ intersect at the point designated P∗.The problem is to show that the points
P and P∗are the same. Figures 6-4(b) and 6-4(c) illustrate that for suitable scalars
k, , m, n, the points P and P∗determine the vectors equations
A+α=mβ and B+nγ =kβ . (6 .6)
In these equations the scalars k, , m, n are unknowns to be determined. Use the
set of equations (6.5), to solve for the vectors α, β, γ in terms of the vectors A and
B and show
α=^12 B−A β=^12 (A+B) γ =^12 A−B. (6 .7)
These equations can now be substituted into the equations (6.6) to yield, after some
simplification, the equations
(1 −−
m
2
)A= (
m
2
−
2
)B and (
k
2
−
n
2
)A= (1 −n−
k
2
)B.
Since the vectors A and B are linearly independent (noncolinear), the scalar coef-
ficients in the above equation must equal zero, because if these scalar coefficients
were not zero, then the vectors A and B would be linearly dependent (colinear)
