Chapter 13.9, Problem 3WE

To determine

To prove: median of a trapezoid is parallel to its bases.

Explanation of Solution

Given information:

  $ABCD$ is a rhombus whose vertices are $A(0,0),B(a,o),C(a+b,c),D(b,c)$ . Joining the diagonals $AC$ and $DB$ .

  

Calculation:

For a rhombus length of opposite sides are equal. That is,

  $\begin{array}{l}\overline{AB}=a\\ \overline{AD}=\sqrt{b^2+c^2}\\ \overline{AB}=\overline{AD}\\ \Rightarrow a^2=b^2+c^2\mathrm{.................}(i)\end{array}$

Now if,

  $\begin{array}{l}\text{Slope of }\overline{AC}=\frac{c-0}{a+b-0}=\frac{c}{a+b}=m_1\\ \text{Slope of }\overline{DB}=\frac{c-0}{b-a}=\frac{c}{b-a}=m_2\end{array}$

Then if both diagonal are perpendicular then it is sufficient to show : $m_1{m}_2=-1$ , so that,

  $m_1{m}_2=\frac{c}{a+b}\times \frac{c}{b-a}=\frac{c^2}{b^2-a^2}=\frac{c^2}{-c^2}=-1\text{ (From (i))}$

Hence it is proved that both diagonals are perpendicular to each other.