Like the adjoining figure, draw two circles with centres C and D which intersect each other at the points A and B . Draw a straight line through A which intersects the circle C at the point P and the circle with centre D at the point Q. Prove that (i) `angle PBQ= angle CAD (ii) angle BPC= angle BQD`.

Explanation

In the circle with centre C,the central angle produced by the chrod `AP angle ACP` and angle in circle = `angle ABP`. By theorem-34,
`angle ACP= 2 angle APB or, angle ABP=(1)/(2) angle ACP......(1)`

Again, in the circle which centre at d, the central angle produced by the chord `AQ=angle ADQ` and angle in circle `angle ABQ`.
`:.` by theorem-34 `angle ADQ= 2 angle ABQ or, angle ABQ=(1)/(2) angle ADQ .....(2)`
then, adding (1) and (2) we get,
`angle ABP+ angleABQ=(1)/(2) angle ACP+(1)/(2)angle ADQ`
or, `angle PBQ=(1)/(2) angle ACP+ angle ADQ`
or, `angle PBQ=(1)/(2)( angle ACP+ angle ADQ)`
`=(1)/(2) (180^(@)-2 angle PAC+ 180^(@)-2 angle QAD)`
`[ :. angle ACP+ angle PAC+ angle PAC =180^(@)`
or, `angleACOP+ angle PAC+ anglePAC=180^(@)`
`or, ACP=180^(@)- 2 angle PAC`
similiarly, `angle ADQ=180^(@)-2 angle QAD`]
or, `angle PBQ=(1)/(2)[ 360^(@)-2 ( angle PAC+ angle QAD)]=(1)/(2) [ 360^(@)-2(180^(@)-angleCAD)]`
`=(1)/(2)[ 360^(@)-360^(@)+2 angle ACD]= (1)/(2) xx angle CAD= angle CAD`
`:. angle PBQ= angle CAD` [ proved (i)]
Now, let us join B,C and B,D
`:. CP= CB ( :.` radii of same circle), `:. angle BPC= angle PBC......(3)`
`:. BD= DQ ( :.` radii of same circle) `:. angle DBQ= angle DQB.......(4)`
Now, from (i) we get,`anglePBQ= angle CAD`.
`or, angle PBC+ angle QBA= angle CAB+angle DAB`
`or, angle PBC+ angle CBA+ angle DAB`
or, `angle PBC+ angle CBA+ angle DBA- angle DBQ= angle CAB+ angle DAB`
or, `angle PBC+ angle CAB+ angle DAB- angle DBQ= angle CAB+ angle DAB`
`[ :. CB= CA, :. angle CBA= CAB`
`:. DB=DA, :. angle DBA= angle DAB]`
or, `angle PBC- angleDBQ=0 or, angle PBC and angle DBQ= angle BQD]`
`:. angle BPC= angle BQD` [ Proved (ii)]
Hence `(i) anglePBQ= angle CAD and (ii) angle BPC= angle BQD` (Proved)