Chapter 15, Problem 39CR

a.

To determine

To prove: that ABCD is a kite.

Answer to Problem 39CR

ABCD is a kite..

Explanation of Solution

Given:

The given figure is shown below triangle.

  

Calculation:

Let us consider ABCD is a kite. Then,

  $\begin{array}{l}CD\cong CB\left\{\text{properties of the kite}\right\}\\ DA\cong AB\left\{\text{properties of the kite}\right\}\end{array}$

Now, find the length of the sides of the triangle by using the distance formula. Then,

  $\begin{array}{l}CD=\sqrt{{\left(x_2-x_1\right)}^2+{\left(y_2-y_1\right)}^2}\\ =\sqrt{{\left(0-\left(-2b\right)\right)}^2+{\left(2a-2c\right)}^2}\\ =\sqrt{4b^2+{\left(2a-2c\right)}^2}\end{array}$

  $\begin{array}{l}CB=\sqrt{{\left(x_2-x_1\right)}^2+{\left(y_2-y_1\right)}^2}\\ =\sqrt{{\left(0-\left(-2b\right)\right)}^2+{\left(2a-2c\right)}^2}\\ =\sqrt{4b^2+{\left(2a-2c\right)}^2}\end{array}$

  $\begin{array}{l}DA=\sqrt{{\left(x_2-x_1\right)}^2+{\left(y_2-y_1\right)}^2}\\ =\sqrt{{\left(0-\left(-2b\right)\right)}^2+{\left(0-2c\right)}^2}\\ =\sqrt{4b^2+4c^2}\end{array}$

  $\begin{array}{l}AB=\sqrt{{\left(x_2-x_1\right)}^2+{\left(y_2-y_1\right)}^2}\\ =\sqrt{{\left(0-\left(-2b\right)\right)}^2+{\left(0-2c\right)}^2}\\ =\sqrt{4b^2+4c^2}\end{array}$

Since, $CD\cong CB,DA\cong AB$

Therefore, ABCD is a kite.

b.

To determine

To prove: that the figure formed by joining consecutive mid-points of the sides of ABCD is a rectangle.

Answer to Problem 39CR

  $C\left(-2,5\right)$ .

Explanation of Solution

Given:

The given figure formed by joining consecutive mid-points of the sides of ABCD is shown below:

  

Calculation:

Let $Q,R,S,T$ be the mis-points of the of $\overline{DA},\overline{DC},\overline{CB},\overline{BA}$ respectively.

From the mid-point formula,

  $\begin{array}{l}Q=\left(-b,c\right)\\ R=\left(-b,a+c\right)\\ S=\left(b,a+c\right)\\ T=\left(b,c\right)\end{array}$

From the slope formula, $m\overline{QR}\text{ and }m\overline{ST}$ are undefined, and thus are parallel, vertical lines $m\overline{QR}\text{ and }m\overline{ST}$ are both 0, and thus are parallel, horizontal lines.

Thus, $QRST$ is a rectangle.