Chapter 10.1, Problem 20PSC

To determine

To prove: that the segment $WQ$ is congruence $ZR$ if $WX\cong YZ.$

Answer to Problem 20PSC

  $WQ\cong ZR$ .

Explanation of Solution

Given:

  

  $\begin{array}{l}WX=YZ.\\ \text{Drop }PN\perp WZ.\end{array}$

Concept used:

Perpendicular from the center to the chord bisects the chords.

Condition of congruence triangle:

Side angle side $\left(SAS\right)$ : two side and one angle of one triangle is equivalent another triangle.

Side-side-side $\left(SSS\right)$ : all three side of the one triangle is equivalent to another triangle.

Angle side angle $\left(ASA\right)$ : two angle and side included between the angles of one triangle is equivalent to another two angle and side included between the angles of triangle.

Angle-angle side $\left(AAS\right)$:

two angle and non-included side between the angles of one triangle is equivalent to another two angle and non-included side between the angles of triangle.

Right angle hypotenuse side $\left(RHS\right)$: Hypotenuse and side of a right-angled triangle is equivalent to the hypotenuse and side of the second right angled triangle.

Calculation:

According to the given:

  $\begin{array}{l}WX=YZ.\\ \text{Drop }PN\perp WZ.\end{array}$

Perpendicular from the center to the chord bisects the chords.

  $\begin{array}{l}XN=NY.\\ \Rightarrow XN+WX=NY+YZ.\\ \Rightarrow WN=NZ.\end{array}$

Consider $\Delta WNP\text{ and }\Delta ZNP$ :

Here by $SAS$ congruence criterion $\Delta WNP$ congruent to $\Delta ZNP$ .

Since, congruent parts of congruent triangles are congruent.

  $\begin{array}{l}WP=ZP.\\ WQ+QP=ZP=ZR+PR.\end{array}$

But, $PQ=PR$ as the radii of the circle is:

Hence, $WQ=ZR$ .