Chapter 6.5, Problem 36PPS

To determine

To prove: The given theorem.

Explanation of Solution

Given information:

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

The diagram:

  

Formula used:

The opposite angles of a parallelogram are congruent.

The sides of a rhombus are congruent.

ASA congruence:

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

Consider that the diagonal WY bisects the opposite angles ZWX and ZYX.

So, $\angle 1\cong \angle 2$ and $\angle 3\cong \angle 4$

Now, in triangles WXY and WZY;

  $\angle 1\cong \angle 2$

WY is the common side

  $\angle 3\cong \angle 4$

So, by ASA congruence;

  $\Delta WXY\cong \Delta WZY$

Since, $\Delta WXY$ and $\Delta WZY$ are congruent;

So, $WZ\cong ZY\cong YX\cong XW$

It is seen that all the sides of the parallelogram are congruent.

Hence, the theorem is proved.