Chapter 5.7, Problem 8PSA

To determine

To prove: That the parallelogram $TWXY$ is a rhombus.

Explanation of Solution

Given Information:

  $YTWX$ is a parallelogram

  $\begin{array}{l}YP\perp TW\\ ZW\perp TY\\ TP\cong TZ\end{array}$

Formula used:

Correspondingparts of congruent triangles are congruent.

Proof:

It is given that $YTWX$ is a parallelogram.

We know that the opposite sides of a parallelogram are equal, so we have

  $YX=TWandYT=XW\mathrm{...}\left(\text{i}\right)$

In $\Delta YTP\text{and}\Delta WTZ,$

  $\angle YTP=\angle WTZ$ … (Common angle)

  $\angle TPY=\angle WZT$ … (Both right angles)

  $TP=TZ$ … (Given)

  $\therefore \Delta YTP\cong \Delta WTZ$ … (By ASA congruency rule)

As we know that the corresponding parts of congruent triangles are congruent

  $\Rightarrow YT=WT$

From (i), we get

  $XW=YT=WT=YX$

Thus, all the sides of a parallelogram are equal.

Hence, $YTWX$ is a rhombus.