Two chords AB and CD of a circle with centre O intersect each other at the point P. prove that `AOD+ angle BOC = 2 angle PBC`. If `angle BOC` are supplementary to each other, then prove that the two chords are perpendicular to each other.

Explanation

In the circle with centre at O, two chords AB and CD intersect each other at P.
To prove : We have prove that `angle AOB+ angle BOC=2 angle BPC and AB bot CD` when `angle AOD+ angle BOC=180^(@)`
Construction : Let us join `O,A, O,B, C,O D and B,D`
Proof : The central angle produced by the are`overset(frown)(AD)=angle AOD` and angle in circle. `= angle ABD`.

`:. angle AOD=2 angle ABD`[ by theorem-34] ......(1)
Now, adding (1) and (2) we get,
`angleAOD+ angleBOC= 2 angle ABD+ 2 angle BDC=2 (angle ABD+angleBDC).....(3)`
But by producing DP of `Delta PBD` upto C, the produced external `angle BPC=` intrnally oppositie `(anglePBD+ angle BDP)` = internally opposite `(angle ABD+ angle BDC) [ :. ` they are angles]
`:.angle ABD+angle BDC= angle BPC`
`:.` from (3) we get , `angle angleAOD+ angleBOC=2 angle BPC`
`:. AOD+ angleBOC=2 angle BPC` (Proved)
Now, if `angle AOD+ angleBOC= 180^(@)`, then we get,
` 180^(@)=2 angle BPC or, angle BPC=(180^(@))/(2)=90^(@)`
Hence `AB bot CD[ :. angle BPC 90^(@)]`