In two circles , one circle passes through the centre O of the other circle and they intersect each other at the points A and B . A straight line passing through A intersect the circle passing through O at the point P and the circle with centre at O at the point R . BY joining P , B and R , B prove that PR= PB.

Explanation

Given : Let the circle with centre at C passes through the centre of other circle with centre at O The two circles intersect each other at the points A and B . A straight line PAR intersects the circle passing through O at the point P and also intersect the circle with centre at O at the point R . The points P , B and B, R are joined .
To prove : PR=PB
Constrution : Let us join O ,A, O, B and O, R
Proof : In the circle with centre at O ,OB = OR [ `because ` radii of same circle ]
`therefore angleOBP=angleORB`.........(1)
Again , PAOB is a cyclic quadrilateral,
`thereforeanglePAO+anglePBO=180^(@)`.........(2)
Now , `anglePAO+angleOAR=1" straight angle"=180^(@)` .......(3)
Then from (2) and (3) we get , `anglePAQ+anglePBO=anglePAO+angleOAR`
or, `anglePBO=angleOAR" or " ,anglePBO=angleORA[becauseOA=OR,becauseangleOAR=angleORA`..........(4)
Now , `anglePBR=anglePBO+angleOBR` [as per figure]
`angleORA+angleORB` [from(4)and (1)]
`anglePRB`
`therefore"in" DeltaPBR,anglePBR=anglePRB,thereforePR=PB`