Two chords intersect at the point P. Prove that....

[abhirup_ab]

Two chords AB and CD of a circle with centre O intersect at the point P. Prove:
<AOD + <BOC = 2<BPC

 

[Glinkx]

Two chords AB and CD of a circle with centre O when produced intersect at the point P. Prove:

1) <AOC - <BOD = 2<BPC
2) <AOD + <BOC = 2<BPC

Are you sure you typed this right? If I add left parts of both statements I get 2<AOB while right parts give 4<BPC. Then I'm free to move chord CD as I pleased changing the value of <BPC while <AOB will stay constant as it does not depend on the position of chord CD. This contradicts the equal sign in the statements above.