Chapter 6.3, Problem 38PPS

To determine

Prove that the segments joining the midpoints of the sides of any quadrilateral form a parallelogram.

Explanation of Solution

Given :

Calculation:

Locate the quadrilateral on the coordinate plane and assume the coordinates of the vertices.

Since A , B , C , and D are midpoints of RS, ST , TU , and UR respectively.

So, their coordinates are :

$A\equiv \left(\frac{0+2a}{2},\frac{0+2f}{2}\right)=(a,f)$

$B\equiv \left(\frac{2a+2d}{2},\frac{2f+2b}{2}\right)=(a+d,f+b)$

$C\equiv \left(\frac{2d+2c}{2},\frac{2b+0}{2}\right)=(d+c,b)$

$D\equiv \left(\frac{0+2c}{2},\frac{0+0}{2}\right)=(c,0)$

Slope of $\overline{AD}\text{ and }\overline{BC}$ :

$m_{AD}=\frac{0-f}{c-a}=\frac{f}{a-c}$

$m_{BC}=\frac{b-(b+f)}{(c+d)-(a+d)}=\frac{-f}{-(a-c)}=\frac{f}{a-c}$

Since , the slope of $\overline{AD}\text{ and }\overline{BC}$ are equal , so $\overline{AD}\parallel \overline{BC}$ .

Now, the lengths of $\overline{AD}\text{ and }\overline{BC}$ :

$\begin{array}{l}AD=\sqrt{{\left(a-c\right)}^2+{\left(f-0\right)}^2}=\sqrt{{\left(a-c\right)}^2+f^2}\\ BC=\sqrt{{\left[\left(c+d\right)-\left(a+d\right)\right]}^2+{\left[b-\left(b+f\right)\right]}^2}=\sqrt{{\left(a-c\right)}^2+f^2}\end{array}$

So, $\overline{AD}\cong \overline{BC}\mathrm{................}[\text{definition of congruency]}$

So, $\overline{AD}\parallel \overline{BC}$ and $\overline{AD}\cong \overline{BC}$ .

Hence , by Theorem 6.12 , ABCD is parallelogram.

Hence proved.