`Delta ABC` is inscirbed in a circle, the bisector of the angles `angle BAC, angle ABC and angle ACB` intersect at the point X,Y,Zon the circle repectively. Prove that in `Delta XYZ , angle YXZ=90^(@)-(1)/(2) angle BAC`

Explanation

`Delta ABC` is inscribed in the circle with centre at O. the bisectors of `angle BAC, angle ABC and angle ACB` intrsect the circle at the points, X Y and Z respectively. Let us join the point X, Y, Y and, X.
To prove `angle YXZ==90^(@)-(1)/(2) angleBAC`
Proof : `angle AYZ and angle ACZ` are two angles in circle produced by the are AZ,
`:. angle AXZ= angle ACZ....... (1)` [ by theorem-35]
Similiarly, `angleAXY and angle ABY` are tow angles in circle produced by the are AY.
`:. angle AXY= angle ABY......(2)` [ by theorem -35]
Now, adding (1) and (2) we get
`angle AXZ+ angle AXY= angle ACZ+ angle ABY`
or, `angle YXZ=(1)/(2) angle C+(1)/(2) angleB=(1)/(2)( angle C+ angleB)`
`=(1)/(2) (180^(@)-angleA)[ :. angleA+ angle B+angleC=180^(@)]`
` =90^(@)-(1)/(2) angleA=90^(@)-(1)/(2) angleBAC`
Hence `angle YXZ=90^(@)-(1)/(2)`