Chapter 6.5, Problem 34PPS

To determine

To prove: The given theorem.

Explanation of Solution

Given information:

The theorem: If a parallelogram is a rhombus, then each diagonal bisects a pair of opposite angles.

The diagram:

  

Formula used:

The opposite angles of a parallelogram are congruent.

The sides of a rhombus are congruent.

SAS congruence:

If two pairs of corresponding sides and one pair of corresponding angles of two triangles are congruent then the two angles are congruent.

For the given parallelogram NPRQ,

It is known that the opposite angles of a parallelogram are congruent.

So, $\angle PNR\cong \angle PQR$

And, $\angle NRQ\cong \angle NPQ$

Also, it is given that $NPRQ$ are rhombus.

It is known that the sides of a rhombus are congruent.

So, $NR\cong RQ\cong QP\cong PN$

Now, for the triangles PNR and PQR;

  $\angle PNR\cong \angle PQR$

  $NR\cong QR$

  $PN\cong QP$

Thus, by SAS congruence $\Delta PNR\cong \Delta PQR$

Since, the triangle PNR and PQR are congruent;

So, the corresponding sides and angles of the above triangles are equal.

Thus, $\angle 1=\angle 2$ ,

  $\angle 3=\angle 4$

This means the diagonal PR bisects the opposite angles $\angle NPQ$ and $\angle NRQ$ .

Similarly, we see that NRQ and NPQ will be congruent.

So, $\angle 5=\angle 6$ ,

  $\angle 7=\angle 8$

This means the diagonal NQ bisects the opposite angles $\angle PQR$ and $\angle PNR$ .

Thus, if a parallelogram is a rhombus, then each diagonal bisects a pair of opposite angles.

Hence, the theorem is proved.