Q. 62 - Calculus

Question

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

It is proved 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 1: Given

A triangle $ABC$, $D \mathrm{and} E$ is the mid point of sides $AB \mathrm{and} AC$ respectively.

Step 2: Proof

From figure,

$AB=\overrightarrow{c}, BC=\overrightarrow{a,} AC=\overrightarrow{b}$

In $∆ABC,$

$$\begin{gathered} \overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}=0 \\ \overrightarrow{a}=-(\overrightarrow{b}+\overrightarrow{c}) ...\left(1\right) \\ \mathrm{In} ∆\mathrm{ADE}, \\ \frac{1}{2}\overrightarrow{c}+\overrightarrow{d}+\frac{1}{2}\overrightarrow{b}=0 \\ \overrightarrow{d}=-\frac{1}{2}\left(\overrightarrow{b}+\overrightarrow{c}\right) \\ \overrightarrow{d}=\frac{1}{2}\overrightarrow{a } (\mathrm{From} \mathrm{equation} 1) \end{gathered}$$

Thus, 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. 

Hence, proved.