If two chords AB and CD of a circle with centre, O, when producd intersect each other at the point P, prove that `angle AOC- angle BOD=2 angle BPC`.

Explanation

Let AB and CD be two chord of the circle with centre at O. produced AB and CD intersect at P.
To prove : We have that ` angle AOC - angle BOD= 2 angle BPC`
Construction : Let us join B,C
Proof : the central angle produced by the arc `overset(frown)(AC)= angle AOC` and angle in circle `= angle ABC`
`:. angle AOC=2 angle ABC ........ (i)`[ by theoremm-34]
Now, substracting (2) from (1) we get,
`angle AOC- angle BOD=-2 angle BCD.......(3)`
Again, by producing PB of `Delta BPC` we, get external `angle ABC`.
`or, angle ABC`=internally oppoisite `( angle BPC+ angle BCP)`
or, ` angle BAC= angle BPC+ angle BCD. or, 2 angle ABC=2 angle BCD`[ multiplying by ]
`or, 2 angle ABC-2 angle BCD=2 angle BPC`
`:.` frm (3) we get, ` angle AOC- angle BOD=2 angle BPC`
Hence, ` angle AOC- angle BOD=2 angle BPC`