Chapter 13.9, Problem 10WE

To determine

To prove: the medians of a triangle intersect in a point called centroid, that is two thirds of the distance from each vertex to the midpoint of the opposite side.

Explanation of Solution

Given information:

A parallelogram $▱ABCD$ given below,

  

Calculation:

We solve the problem by vector method. In the taken figure,

  $\begin{array}{l}\overrightarrow{AC}=\overrightarrow{AB}+\overrightarrow{BC}\\ \overrightarrow{BD}=\overrightarrow{BC}-\overrightarrow{AB}\end{array}$

Since in parallelogram $\left|\overrightarrow{AD}\right|=\left|\overrightarrow{BC}\right|,\left|\overrightarrow{AB}\right|=\left|\overrightarrow{DC}\right|$ , we have to show ,

  ${\left|\overrightarrow{AC}\right|}^2+{\left|\overrightarrow{BD}\right|}^2=2{\left|\overrightarrow{BC}\right|}^2+2{\left|\overrightarrow{AB}\right|}^2$

Now consider,

  ${\left|\overrightarrow{AC}\right|}^2+{\left|\overrightarrow{BD}\right|}^2={\left|\overrightarrow{BC}+\overrightarrow{AB}\right|}^2+{\left|\overrightarrow{BC}-\overrightarrow{AB}\right|}^2=2\left({\left|\overrightarrow{BC}\right|}^2+{\left|\overrightarrow{AB}\right|}^2\right)=2{\left|\overrightarrow{BC}\right|}^2+2{\left|\overrightarrow{AB}\right|}^2$

Since ${(a+b)}^2+{(a-b)}^2=2(a^2+b^2)$ .

Hence proved.