Chapter 9.1, Problem 13P

To determine

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

Explanation of Solution

Given information: The points are $A\left(0,0\right),B\left(2a,0\right),C\left(2b,2c\right)$ and $D\left(2d,2e\right)$ .

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.

  $\begin{array}{l}\Rightarrow AB=\left(\frac{0+2a}{2},\frac{0+0}{2}\right)\\ =\left(a,0\right)=w\\ \Rightarrow BC=\left(\frac{2a+2b}{2},\frac{0+2c}{2}\right)\\ =\left(a+b,c\right)=x\\ \Rightarrow CD=\left(\frac{2b+2d}{2},\frac{2c+2e}{2}\right)\\ =\left(b+d,c+e\right)=y\\ \Rightarrow DA=\left(\frac{2d+0}{2},\frac{2e+0}{2}\right)\\ =\left(d,e\right)=z\end{array}$

Get the lengths of the sides of the quadrilateral,

  $\begin{array}{l}\Rightarrow WX=\sqrt{{\left(a-a-b\right)}^2+{\left(0-c\right)}^2}\\ =\sqrt{b^2+c^2}\\ \Rightarrow XY=\sqrt{{\left(a+b-b-d\right)}^2+{\left(c-c-e\right)}^2}\\ =\sqrt{a^2+d^2+e^2-2ad}\end{array}$

Similarly,

  $\begin{array}{l}\Rightarrow YZ=\sqrt{{\left(b+d-d\right)}^2+{\left(c+c-e\right)}^2}\\ =\sqrt{b^2+c^2}\\ \Rightarrow ZW=\sqrt{{\left(d-a\right)}^2+{\left(c-0\right)}^2}\\ =\sqrt{a^2+b^2+e^2-2ad}\end{array}$

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

  $b^2+c^2+a^2+d^2+e^2-2ad+b^2+c^2+a^2+d^2+e^2-2ad=2a^2+2b^2+2c^2+2d^2+2e^2-4ad$

Now, get the lengths of the diagonals of the quadrilateral,

  $\begin{array}{l}\Rightarrow WY=\sqrt{{\left(a-b-d\right)}^2+{\left(0-c-e\right)}^2}\\ =\sqrt{a^2+b^2+c^2+d^2+e^2-2ab-2ad+2bd+2ce}\\ \Rightarrow XZ=\sqrt{{\left(a+b-d\right)}^2+{\left(c-e\right)}^2}\\ =\sqrt{a^2+b^2+c^2+d^2+e^2+2ab-2ad+2bd-2ce}\end{array}$

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

  $\begin{array}{l}{a}^2+b^2+c^2+d^2+e^2-2ab-2ad+2bd+2ce+a^2+b^2+c^2+d^2+e^2+2ab-2ad+2bd-2ce\\ =2a^2+2b^2+2c^2+2d^2+2e^2-4ad\end{array}$

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.