Chapter 10, Problem 10RP

To determine

To prove : That $\stackrel{⌢}{WX}\cong \stackrel{⌢}{YZ}$

Explanation of Solution

Given information :

We know that, $OP\perp WX,OQ\perp YZ.$

  $\Delta OPQ$ is an isosceles triangle.

  

Construction :

OX, OW, OY, OZ are joined.

Calculation :

In $\Delta OPX$ and $\Delta OQY(\because OP\perp WX\text{ and }OQ\perp YZ)$

OX = OY (radius of the circle)

OP = OQ ( $\because \Delta OPQ$ is an isosceles triangle)

  $\angle OPX=\angle OQY$

  $\therefore \Delta OPX\cong \Delta OQY$ (SAS congruency)

  $\therefore \angle XOP=\angle YOQ$ (By C.P.C.T.) …(1)

Similarly, we can prove that

  $\Delta OPW\cong \Delta OPX$

And by C.P.C.T.

  $\angle XOP=\angle WOP$    …..(2)

And

  $\Delta YOQ\cong \Delta OQZ(\therefore \angle YOP=\angle ZOQ)$ ….(3)

  $\Delta QOZ\cong \Delta AOW(\therefore \angle QOZ=\angle POW)$ ….(4)

From equations (1), (2), (3) and (4), we get

  $\angle XOW=\angle YOZ$

Since, the angle extended by $\stackrel{⌢}{XW}$ and $\stackrel{⌢}{ZY}$ are congruent. So, the arc itself are congruent.

i.e. $\stackrel{⌢}{XW}\cong \stackrel{⌢}{ZY}$