Question

62. Use a vector argument to prove that the segment connecting the midpoints of two sides of a triangle is parallel to the third side of the triangle and half of its length.

Short answer

• A triangle midsegment is a segment that connects the midpoints of two triangle sides.
• Any two sides of a triangle's midpoints are parallel to the third side.

Step-by-step solution

Introduction

Consider the triangle$ABC$with vertices$a,b$and $c$

The objective is to use vector argument to prove that segment connecting mid-points of two sides of a triangle is parallel to the third side of the triangle and half of its length.

Consider the triangle $ABC$with vertices representing the vectors $\mathbf{a},\mathbf{b}$, and $\mathbf{c}$as shown below.

Triangle $ABC$

Given Information.

The mid-point between the vectors $\mathbf{a}$and $\mathbf{b}$is:

$\frac{\mathbf{a}+\mathbf{b}}{2}$

The mid-point between the vectors $\mathbf{a}$and $c$is:

$\frac{a+c}{2}$

Now, compare the line segment from $B$to $C$and from $D$to $E$.

The vector from $B$to $C$is:

$\overrightarrow{BC}=\mathbf{c}-\mathbf{b}$

Explanation (part a)

The vector from $DtoE$is:

$\overrightarrow{DE}=\raisebox{1ex}{$a+c$}\!\left/ \!\raisebox{-1ex}{$2$}\right.\raisebox{1ex}{$-a+b$}\!\left/ \!\raisebox{-1ex}{$2$}\right.$

$$\begin{gathered} =\raisebox{1ex}{$a+c-a-b$}\!\left/ \!\raisebox{-1ex}{$2$}\right. \\ =\raisebox{1ex}{$c-b$}\!\left/ \!\raisebox{-1ex}{$2$}\right. \end{gathered}$$

Explanation (part b)

The vector $\overrightarrow{DE}$is half of the vector $\overrightarrow{BC}$and is in the direction of the vector $\overrightarrow{BC}$. Therefore, the vector $\overrightarrow{DE}$ is parallel to the vector$\overrightarrow{BC}$

Hence, segment connecting mid-points of two sides of a triangle is parallel to the third side of the triangle and half of its length.