Q.10.2 - Calculus

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.

Step-by-Step Solution

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 1: Introduction

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

Triangle $A B C$

Step 2: 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:

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

Step 3: Explanation (part a)

The vector from $D to E$ is:

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

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

Step 3: 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.