Two circles whose centres are `O` and `O '` intersect at `Pdot` Through `P ,` a line `l` parallel to `O O '` intersecting the circles at `C\ a n d\ D` is drawn. Prove that `C D=2\ O O '`

Explanation

Draw perpendiculars `O'A \and\ O'B' \to\ `CD \from\ `O \and\ O′`,respectively.
Since `OA⊥CD, OA` bisects the chord CP. (Perpendicular from the centre of the chord bisects the chord)
`AP=1/2CP,CP=2A`- - - - - (i)
Similarly since `O′B⊥PD`, we must have `BP=1/2PD or PD=2BP`- - - - - (ii)
Now, `CD=CP+DP`
`=>CD=2AP+2BP=2(AP+BP)`(from (i) and (ii))
`=>CD=2(AB)`- - - - - (iii)
In quadrilateral `ABO′O, OO′∥AB \and\ /_OAB=/_O'BA=90^0`
Thus, `ABO'O` is a rectangle.
`AB=OO', \Hence\ CD=2AB=2OO′`