In the given figure, ` AB` is a diameter of a circle `C(O,r)` and chord `CD` `=` radius `OC`. If `AC` and `BD` when produced meet at `P`, prove that `/_ APB` is constant.

Explanation

Join `OD` and `BC.`
Now, `CD =OC =OD ` [ each equal to `r` ]
`implies Delta OCD ` is equilateral
`implies /_ COD = 60^(@)` ….(i)
Also, `/_ ACB = 90^(@)` [angle in a semicircle ] …..(ii)
And, `/_ CBD = (1)/(2) xx /_ COD = ((1)/(2) xx60^(@)) = 30^(@) ` ....(iii)
`[ :'` angle made by `hat (CD)` at the centre
`= 2 xx ` angle at any point on its remaining part ] .
Now , side `PC` of `Delta BPC` is produced to `A.`
`:. `exterior angle `=` sumof interior opposite angles.
`implies/_ACB = /_ APB + /_ CBD`
`implies 90^(@) = /_ APB +30^(@) ` [ using (ii) and (iii) ]
`implies /_ APB = 60^(@) =` constant.
Hence, `/_ APB `is constant.