Chapter 13.9, Problem 5WE

To determine

To prove: that the segment joining the midpoints of the diagonals of a trapezoid is parallel to the bases and has a length equal to half the difference of the lengths of the bases.

Explanation of Solution

Given information:

  $MNOP$ is a trapezoid with vertices $M(d,c),N(b,c),O(0,0),P(a,0)$ .

  

Calculation:

Let us consider the mid points $M_1,M_2$ of the sides then,

  $\begin{array}{l}{M}_1=\left(\frac{b+0}{2},\frac{c+0}{2}\right)=\left(\frac{b}{2},\frac{c}{2}\right)\\ M_2=\left(\frac{a+d}{2},\frac{c+0}{2}\right)=\left(\frac{a+d}{2},\frac{c}{2}\right)\end{array}$

Now $MN\left|\right|OP$

  $\begin{array}{l}\text{Slope of }\overline{M_1{M}_2}=\frac{\frac{c}{2}-\frac{c}{2}}{\frac{a+d}{2}-\frac{b}{2}}=0\\ \text{Slope of }\overline{OP}=\frac{0-0}{a-0}=0\end{array}$

Which implies $\overline{M_1{M}_2}$ is parallel to $\overline{OP}$ .

  $\begin{array}{l}\text{Length of base }\overline{OP}=a\\ \text{Length of base }\overline{MN}=\sqrt{{(d-b)}^2+{(c-c)}^2}=d-b\\ \text{Difference of length of bases =a-b+d}\end{array}$

  $\text{Length of }\overline{M_1{M}_2}\text{ =}\sqrt{{\left(\frac{a+d}{2}-\frac{b}{2}\right)}^2+{\left(\frac{c}{2}-\frac{c}{2}\right)}^2}=\sqrt{{\left(\frac{a+d-b}{2}\right)}^2}=\frac{a+d-b}{2}=\frac{1}{2}(\text{Difference of length of bases})$

Hence proved the result.