Question

Bisectors of angle A ,B,C of a triangle ABC intersect its circumcircle at D,E,F respectively prove that angles of triangle DEF are 90-A/2,90-B/2 and90-C/2

Answer 1

Answer:hard question

Step-by-step explanation:

Answer 2

Answer:

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)

Adding equations (i) and (ii) we have,

∠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)]