Two chords AC and BD of a circle intersect each other at the point O. If two tangents drawn at the points A and B intersect each other at the point P and two tangents drawn at the points C and D intersect at the point Q, prove that `angle P+ angle Q =2 angle BOC. [GP-X]`

Explanation

In the circel with centre O', two chords AC and BD intersect each other at O. Two tangents PAS and PBT drawn at the points A and B respectively intersect each oter at P. Again, the tangents QCN and QDM drawn at C and D respectively intersect each other at Q.
To prove : `angle P + angle Q=2 angle BOC.`
Construction : Let us join O.' A, O', D, A, B and C, D.
Proof : From example 8 above, we get
`P=2 angle BAO' and Q=2 angle DCO'` ltbr `=2 (angle BAQ' + angleDCO')`
`=2 (angle BAC +angle CAO' +angle DCO') [becauseBAO'= angle BAC+angle CAO']`
`=2(angle BDC+angle ACO' +angle DCO')`
`=2 (angle BDC +angle A CD [because angle ACO' +angleDCO'= angle ACD]`
`=2 (angle ODC +angle OCD ([because angle BDC=angle OBC, angle ACD=angle OCD]`
`=2 angleBOC[because "in" DeltaCOD` externla `angle BOC=` internally opposite` (angle ODC+angle OCD]`
Hence `angleP +angle Q =2 angle BOC` (Proved)