Here, ABC is inscribed in a circle with center O and the bisectors of ∠A, ∠B and ∠C intersect the circumcircle at D, E and F respectively.

Now, join DE, EF and FD

As angles in the same segment are equal, so,

∠FDA = ∠FCA ...........(i)

∠FDA = ∠EBA ...........(i)

By adding equations (i) and (ii) we get,

∠FDA+∠EDA = ∠FCA+∠EBA

Or, ∠FDE = ∠FCA+∠EBA = (1/2)∠C + (1/2)∠B

We know, ∠A + ∠B + ∠C = 180°

So, ∠FDE = (1/2)[∠C+∠B] = (1/2)[180°-∠A]

∠FDE = [90-(∠A/2)]

In a similar way,

∠FED = [90° -(∠B/2)]

∠EFD = [90° -(∠C/2)]