The internal and external bisectors of the vertical angle of a triangle intersect the circumcircle of the triangle at the points P and Q. Prove that PQ is a diameter of the circle.

Answer

page 292

Explanation

AP is the internal bisector of the vertical angle `angle A` of the `Delta ABC` and AE is the external bisector of the vertical `angle A "of" Delta ABC` and they intersect the circumcircle of the `Delta ABC` at the points P and Q respectively.
To prove : PQ is a diameter of the circle.
Proof : `angle BAC+ angle CAD=1` straight angle `180^(@) alpha, or (1)/(2) angle BAC+ (1)/(2) angleCAD=90^(@)` [ dividing by 2]
or `angle PAC+ angle CAE=90^(@)[ :. AP` is the internal bisector of `angleA ` and E is the external bisector of `angleA`]
or, `angle PAQ= 90^(@)`
`:. angle PAQ` is a semicircular angle
Hence PQ is a diameter of the circle.