Chapter 5.1, Problem 17CE

a.

To determine

To observe what is true about the quadrilateral $ABCD$ .

Answer to Problem 17CE

Quadrilateral $ABCD$ is a parallelogram

Explanation of Solution

Given:

Two segments $AC$ and $BD$ , that bisect each other at $O$ .

  $\begin{array}{l}\Rightarrow OB=OD\\ OA=OC\end{array}$

Thus it forms a quadrilateral $ABCD$ .

  

The quadrilateral $ABCD$ appears to be a parallelogram.

Proof:

In $\Delta AOB$ and $\Delta COD$

  $OA=OC$ (Given)

  $OB=OD$ (Given)

  $\angle AOB=\angle COD$ (Vertically opposite angles)

Therefore, $\Delta AOB\cong \Delta COD$ by Side-Angle-Side (SAS) congruency criteria.

  $\Rightarrow AB=CD$ and $\angle OAB=\angle OCD$ (Corresponding parts of congruent triangles are congruent (equal))

Now since $\angle OAB=\angle OCD$ , so $AB$ is parallel to $CD$ (alternate interior angles are equal).

So, $AB=CD$ and $AB$ is parallel to $CD$ .

Hence, quadrilateral $ABCD$ is a parallelogram.

Conclusion:

Therefore, quadrilateral $ABCD$ is a parallelogram.

b.

To determine

To find the theorem whose converse is used in this exercise.

Answer to Problem 17CE

The diagonals of a parallelogram bisect each other.

Explanation of Solution

Given:

Parallelogram $ABCD$ .

  $\Rightarrow AB=CD$ and $AD=CB$

  $AB$ is parallel to $CD$ and $AD$ is parallel to $CB$ .

Diagonals $AC$ and $BD$ intersect at point $O$ .

  

To prove:

  $AO=OC$

  $BO=OD$

Proof:

In $\Delta AOB$ and $\Delta COD$

  $\angle OAB=\angle OCB$

  $\angle AOB=\angle COA$

  $AB=CD$

Therefore, $\Delta AOB\cong \Delta COD$ by Angle-Angle-Side (AAS) congruency criteria.

  $\Rightarrow AO=OC$ and $BO=OD$ (Corresponding parts of congruent triangles are congruent (equal)).

Conclusion:

Therefore, the diagonals of a parallelogram bisect each other.