Chapter 15.3, Problem 22PSC

To determine

To prove: that $\angle YZW>\angle XYW$ as per given condition.

Answer to Problem 22PSC

  $\angle YZW>\angle XYW$ .

Explanation of Solution

Given:

As $\overline{WX}$ is the diameter of circle with center P , and $\overline{YZ}$ is the diameter of a larger concentric circle.

The non-collinear points of the concentric circle are $W,X,Y,\text{and }Z$ .

Calculation:

According to the question, there is a concentric circle whose center is P and the larger diameter is $\overline{YZ}$ and the smaller diameter is $\overline{WX}$ , and $W,X,Y,\text{and }Z$ are the non-collinear points.

Thus, the circle is like this:

  

Now, in triangle $WPZ\text{and }XPY$ ,

  $\begin{array}{l}\angle WPZ=\angle XPY\left\{\text{vertex are equal}\right\}\\ \overline{YP}\cong \overline{PZ}\left\{\text{radius of the circle are equal}\right\}\\ \overline{PX}\cong \overline{WP}\left\{\text{radius of the circle are equal}\right\}\\ \Delta WPZ\cong \Delta \angle XPY\left\{by\text{SAS-Congruence rule}\right\}\\ \therefore \overline{WZ}\cong \overline{XY}\{\text{by C}\text{.P}\text{.CT}}\end{array}$

Now, comparing triangle $WYZ\text{and }XWY$ ,

  $\begin{array}{l}\overline{WY}\cong \overline{WY}\left\{\text{reflexive property}\right\}\\ \overline{WZ}\cong \overline{XY}\left\{above\text{proved}\right\}\\ \overline{YZ}>\overline{WX}\left\{\text{diameter of larger circle is greater}\right\}\\ \therefore \angle YZW>\angle XYW\left\{by\text{the converse of Hinge theorem}\right\}\end{array}$

Thus, $\angle YZW>\angle XYW$ .

Hence, the theorem is proved.