Diagonals of a parallelogram

The use of vectors is very well illustrated by the following rather famous proof that the diagonals of a parallelogram mutually bisect one another.

Figure 13: A parallelogram

Suppose that the quadrilateral ABCD in Fig. 13 is a parallelogram. It follows that the opposite sides of ABCD can be represented by the same vectors, ${\bf a}$ and ${\bf b}$: this merely indicates that these sides are of equal length and are parallel (i.e., they point in the same direction). Note that Fig. 13 illustrates an important point regarding vectors. Although vectors possess both a magnitude (length) and a direction, they possess no intrinsic position information. Thus, since sides $AB$ and $DC$ are parallel and of equal length, they can be represented by the same vector ${\bf a}$, despite the fact that they are in different places on the diagram.

The diagonal $BD$ in Fig. 13 can be represented vectorially as ${\bf d} = {\bf b} - {\bf a}$. Likewise, the diagonal $AC$ can be written ${\bf c} = {\bf a} + {\bf b}$. The displacement ${\bf x}$ (say) of the centroid $X$ from point $A$ can be written in one of two different ways:

$${\bf x} = {\bf a} + \lambda {\bf d},$$ (38)

$${\bf x} = {\bf b} + {\bf a} - \mu {\bf c}.$$ (39)

Equation (38) is interpreted as follows: in order to get from point $A$ to point $X$, first move to point $B$ (along vector ${\bf a}$), then move along diagonal $BD$ (along vector ${\bf d}$) for an unknown fraction $\lambda$ of its length. Equation (39) is interpreted as follows: in order to get from point $A$ to point $X$, first move to point $D$ (along vector ${\bf b}$), then move to point $C$ (along vector ${\bf a}$), finally move along diagonal $CA$ (along vector $-{\bf c}$) for an unknown fraction $\mu$ of its length. Since $X$ represents the same point in Eqs. (38) and (39), we can equate these two expressions to give

$${\bf a} + \lambda ({\bf b} - {\bf a}) = {\bf b} + {\bf a} - \mu ({\bf a} + {\bf b}).$$ (40)

Now vectors ${\bf a}$ and ${\bf b}$ point in different directions, so the only way in which the above expression can be satisfied, in general, is if the coefficients of ${\bf a}$ and ${\bf b}$ match on either side of the equality sign. Thus, equating coefficients of ${\bf a}$ and ${\bf b}$, we obtain

$$1 - \lambda = 1 - \mu,$$ (41)

$$\lambda = 1 -\mu.$$ (42)

It follows that $\lambda=\mu =1/2$. In other words, the centroid $X$ is located at the halfway points of diagonals $BD$ and $AC$: i.e., the diagonals mutually bisect one another.