Chapter 10.1, Problem 4CFU

To determine

Prove that a quadrilateral is a parallelogram.

Answer to Problem 4CFU

  $ABCD$ is a parallelogram.

Explanation of Solution

Given information:

The coordinates of the vertices.

Calculation:

Here, we will consider $A\left(0,0\right),B\left(0,a\right),C\left(a,b\right),D\left(0,b\right)$ :

  

Now, we will consider $case-1$ if both pairs of opposite sides of a quadrilateral are equal then it is a parallelogram, because a rectangle is a parallelogram.

Now, distance between the point $AB=\sqrt{{\left(a-0\right)}^2+{\left(0-0\right)}^2}=a$

Now, distance between the point $CD=\sqrt{{\left(a-0\right)}^2+{\left(b-b\right)}^2}=a$

Therefore,

  $AB=CD$

Now, distance between the point $AD=\sqrt{{\left(0,0\right)}^2+{\left(b-0\right)}^2}=b$

Now, distance between the point $BC=\sqrt{{\left(a-0\right)}^2+{\left(b-0\right)}^2}=b$

Therefore,

  $AD=BC$

Thus, $ABCD$ is a parallelogram.

Now, we will consider $case-2$ if diagonals bisect each other than the quadrilateral is a parallelogram.

Now, mid-point of diagonal $AC=\left(\frac{a+0}{2},\frac{b+0}{2}\right)=\left(\frac{a}{2},\frac{b}{2}\right)$

Now, mid-point of diagonal $BD=\left(\frac{a+0}{2},\frac{0+b}{2}\right)=\left(\frac{a}{2},\frac{b}{2}\right)$

Here, we can see that two diagonals bisect each other. Thus, $ABCD$ is a parallelogram.

Now, we will consider $case-3$ if one pair of opposite sides is parallel and equal then it is a parallelogram.

From $case-1,AB=CD\left[oppositesidesareequal\right]$

Slope of side $AB$ :

  $\begin{array}{l}AB=\frac{y_2-y_1}{x_2-x_1}\\ AB=\frac{0-0}{a-0}=0\end{array}$

Slope of side $CD$ :

  $\begin{array}{l}CD=\frac{y_2-y_1}{x_2-x_1}\\ CD=\frac{b-b}{a-0}=0\end{array}$

Therefore, opposite sides are parallel. Thus, $ABCD$ is a parallelogram.

Now, we will consider $case-4$ if all the four sides of a quadrilateral are equal then it is a parallelogram.

From $case-1$ we have $AB=BC=CD=DA$

Now, provided $a=b$

Thus, $ABCD$ is a parallelogram.

Hence, the results are shown above.