Bisectors of angles A, B and C of a triangle ABC intersect its circumcircle at D, E and F respectively. Prove that the angles of the triangle DEF are 90° - A, 90° - B and 90° - C.

In ΔDEF, ∠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° - ∠A
Similarly, ∠E = 90° - ∠B and ∠F = 90° - ∠C