Chapter 5.4, Problem 6CE

To determine

To prove: that if two consecutive sides of a parallelogram are congruent then it is a rhombus.

Explanation of Solution

Given:

A parallelogram with two congruent consecutive sides.

  

$AB=BC$

Proof:

  $ABCD$ is a given parallelogram.

Now, $AB$ is parallel to $DC$ ,

The distance between two parallel lines is equal.

  $\Rightarrow AD=BC$ ….(i).

Similarly, $AD$ is parallel to $BC$ ,

  $\Rightarrow AB=DC$ ….(ii).

Now, $AB=BC$ ….(iii).

From equation (i), (ii) and (iii),

  $AB=BC=CD=DA$ ….(iv).

Join $AC$ and $DB$ .

Now, the diagonals of a parallelogram bisect each other.

So, $DO=OB$ and $AO=OC$ .

In $\Delta AOD$ and $\Delta AOB$

  $AD=AB$ (proved above)

  $DO=OB$ (diagonals bisect each other)

  $AO=AO$ (common)

  $\therefore \Delta AOD\cong \Delta AOB$ by Side-Side-Side(SSS) congruency criteria.

  $\Rightarrow \angle AOD=\angle AOB$ (corresponding parts of congruent triangles are congruent(equal))

Now,

  $\angle AOD+\angle AOB={180}^0$ (Linear Pair)

  $\angle AOD+\angle AOD={180}^0$ ( $\because \angle AOD=\angle AOB$ )

  $2\angle AOD={180}^0$

  $\Rightarrow \angle AOD={90}^0$

Similarly, $\angle AOD=\angle AOB=\angle DOC=\angle COB={90}^0$ ….(v).

  $\therefore$ by (iv) and (v),

The given quadrilateral is a rhombus.

Hence proved.

Conclusion:

Therefore, the given parallelogram with congruent consecutive sides is a rhombus.