The isosceles triangle ABC is inscribed in the circle with centre at O.If AP be a diamtere passing through A, then show that AP is internal bisector of `angle BPC`.

Explanation

`Delta ABC` is an isosceles triangle of which `AB=AC`. O is the centre of the in which `Delta ABC` is insrbied and AP is dimeter of the circle.
To prove : AP is the internal bisector of `Delta BPC,i.e. angle APB= angleAPC`.
Construction : Let us join the point `O,B,C,C,B,P and C,P`.
Proof In `Delta ABC,AB=AC, :. angle AOB= angleAOC[ :. ` equal segments of a circle prodcued equal front angles at the centre]
or , `2 angle APB=2 angle APC` [ by thorem -34] `or, angle APB= angleAPC`.
Hence AP is the internal bisector of `angle BPC`.