Bisectors of angles A, B and C of a triangle ABC intersect its circumcircle at D, E andF respectively. Prove that the angles of the triangle DEF are `90^@-1/2A`,`90^@-1/2B`and `90^@-1/2C`

Explanation

We see that
In `/_\DEF,`
`/_D=/_EDF`

But `/_EDF=/_EDA+/_FDA` (Angle addition property)
Now,
`/_EDA=/_EBA and /_FDA=/_FCA` (.Angles inscribed in the same arc)
`:./_EDF=/_EBA+/_FCA= 1/2/_B+ 1/2/_C`
[Since `BE` is bisector of `/_B` and `CF` is bisector `/_C`]
`:./_D= /_B+(/_C)/2 ....(1)`
Similarly,
`/_E= (/_C+/_A)/2` and `/_F= (/_A+/_B)/2 ...(2)`
Now,
`/_A+/_B+/_C=180^@` (Angle sum property of triangle)
`:./_B+/_C=180^@ −/_A ...(3)`
Similarly,
`/_C+/_A=180^@−/_B` and `/_A+/_B=180^@−/_C .....(4)`
Substituting eq `(3)` in eq `(1)`, we get
`:./_D=(180^@−/_A)/2`
`/_D=90^@ − 1/2 /_A`
Similarly,
`/_E=90^@− 1/2 /_B` and `/_F=90^@ − 1/2 /_C`
Hence proved.