Chapter 10.2, Problem 3PSA

To determine

To prove: that the $W$ is congruence to $Y$ if $PR$ is perpendicular to $PS$ .

Answer to Problem 3PSA

  $W\cong Y$ .

Explanation of Solution

Given:

  

  $WP=PY=r$ .

Concept used:

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 draw $WP\text{ and }PY$ ,since these lines are radius, then there are equal.

  $WP=PY=r$ then $\angle PWY=\angle PYW$ .

Since, $WP=PY\text{ and }PR=PS\text{}\text{ and }\angle WRP=\angle YSP={90}^{0.}$ .

Then all sides are equal. Therefore, $SSS$ similarity $\Delta WRP=\Delta YSP$ , then the angle $\angle RWP=\angle SYP$ .

That means $\angle RWP+\angle PWY=\angle W=\angle Y=\angle PYS+\angle PYW$ .

Hence, $W\cong Y$ .