Question -
Answer -
Consider the followingdiagram
Here, ABC is inscribedin a circle with center O and the bisectors of ∠A, ∠B and ∠C intersect the circumcircle at D, E and Frespectively.
Now, join DE, EF andFD
As angles in the samesegment 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 = (½)∠C+(½)∠B
We know, ∠A +∠B+∠C = 180°
So, ∠FDE = (½)[∠C+∠B] = (½)[180°-∠A]
∠FDE = [90-(∠A/2)]
In a similar way,
∠FED = [90° -(∠B/2)] °
And,
∠EFD = [90° -(∠C/2)] °