Chapter 9.1, Problem 8P

To determine

To prove: The midpoints of a quadrilateral are the vertices of a parallelogram.

Explanation of Solution

Given information: The points are $A\left(-2,3\right),B\left(2,7\right),C\left(6,1\right)$ and $D\left(0,-3\right)$ .

Formula used: $D=\sqrt{{\left(c-a\right)}^2+{\left(d-b\right)}^2}$ and $\left(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2}\right)$

Proof: Get the midpoints of the given coordinates.

  $\begin{array}{l}AB=\left(\frac{-2+2}{2},\frac{3+7}{2}\right)=\left(0,5\right)\\ BC=\left(\frac{2+6}{2},\frac{7+1}{2}\right)=\left(4,4\right)\\ CD=\left(\frac{6+0}{2},\frac{1-3}{2}\right)=\left(3,-1\right)\\ DA=\left(\frac{0-2}{2},\frac{-3+3}{2}\right)=\left(-1,0\right)\end{array}$

Let $W\left(0,5\right),X\left(4,4\right),Y\left(3,-1\right)$ and $Z\left(-1,0\right)$ .

Get the lengths of the sides of the quadrilateral,

  $\begin{array}{l}\Rightarrow WX=\sqrt{{\left(0-4\right)}^2+{\left(5-4\right)}^2}\\ =\sqrt{16+1}=\sqrt{17}\\ \Rightarrow XY=\sqrt{{\left(4-3\right)}^2+{\left(4+1\right)}^2}\\ =\sqrt{1+25}=\sqrt{26}\end{array}$

Similarly,

  $\begin{array}{l}\Rightarrow YZ=\sqrt{{\left(3+1\right)}^2+{\left(-1-0\right)}^2}\\ =\sqrt{16+1}=\sqrt{17}\\ \Rightarrow ZW=\sqrt{{\left(-1-0\right)}^2+{\left(0-5\right)}^2}\\ =\sqrt{1+25}=\sqrt{26}\end{array}$

Hence, the sum of the square of the sides of quadrilateral WXYZ:

  $17+26+17+26=86$

Now, calculate the length of the diagonals of the quadrilateral WXYZ:

  $\begin{array}{l}\Rightarrow WY=\sqrt{{\left(0-3\right)}^2+{\left(5+1\right)}^2}\\ =\sqrt{9+36}\\ =3\sqrt{5}\\ \Rightarrow XZ=\sqrt{{\left(4+1\right)}^2+{\left(4-0\right)}^2}\\ =\sqrt{25+16}\\ =\sqrt{41}\end{array}$

Similarly the sum of the squares of the diagonals of quadrilateral WXYZ:

  $45+41=86$

Hence, the sum of the squares of the sides of a quadrilateral is equal to the sum of the squares of the lengths of its diagonals.

Therefore, the midpoints of a quadrilateral are the vertices of a parallelogram.

Hence, proved.