Chapter 13.9, Problem 8WE

To determine

To prove: that the quadrilateral formed by joining mid points of the sides of an isosceles trapezoid is a rhombus.

Explanation of Solution

Given information:

Given a isosceles trapezoid $\square ABCD$ .

Calculation:

Let us consider the following figure where $F,G,H,I$ are the mid points of the sides of the trapezoid $\overline{AB},\overline{BC},\overline{CD},\overline{DA}$ respectively.

  

Since $AB\left|\right|CD$ , and $\left|AD\right|=\left|BC\right|$ . $FGHI$ is a rhombus when all sides are equal in length.

The solution will takes two steps-

Step-1 First we have to show $\angle ADC=\angle BCD$ . Consider the following figure where $B$ is extended to $E$ which is parallel to $AD$ .

  

So that, $\left|AD\right|=\left|BE\right|=\left|BC\right|$ . Therefore $\Delta BEC$ is isosceles and,

  $\angle BEC=\angle DCB\mathrm{..................}(i)$

Since $AD\left|\right|BE$ implies, $\angle ADC=\angle BEC$ ……………(ii)

From (i) and (ii) it follows, $\angle ADC=\angle BCD$ .

Step-2 To show that $\left|AC\right|=\left|DB\right|$ considering the following figure by joining two diagonal,

  

And using step-1 and side-angle-side, $\Delta ADC$ is congruent with $\Delta BDC$ .

Finally to show the main result we joining $AC$ and $DB$ as follows together with mid points joining.

  

We can see for diagonal $AC$ in $\Delta ABC$ , $\left|AC\right|=2\left|FG\right|$ , and in $\Delta ADC$ , $\left|AC\right|=2\left|IH\right|$ . Similarly for diagonal $DB$ , $\left|DB\right|=2\left|HG\right|$ .

Since, $\left|AC\right|=\left|DB\right|$ , it follows that,

  $\left|FG\right|=\left|IH\right|=\left|IF\right|=\left|HG\right|$

Hence proved that the quadrilateral formed by joining mid points of the sides of an isosceles trapezoid is a rhombus.