Question

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° – 1/ 2 A, 90° – 1/ 2 B and 90° – 1/ 2 C

Answer 1

Given : ∆ABC and its circumcircle. AD, BE, CF are bisectors of ∠A, ∠B, ∠C respectively.

Construction : Join DE, EF and FD.

Proof : We know that angles in the same segment are equal.

∴ ∠5 = ∠C 2 and ∠6 = ∠B 2 ..(i)

∠1 = ∠A 2 and ∠2 = ∠C 2 ..(ii)

∠4 = ∠A 2 and ∠3 = ∠B 2 ..(iii)

From (i), we have

∠5 + ∠6 = ∠C/ 2 + ∠B/ 2

⇒ ∠D = ∠C/ 2 + ∠B/ 2 ...(iv)

[∵∠5 + ∠6 = ∠D]

But ∠A + ∠B + ∠C = 180°

⇒ ∠B + ∠C = 180° – ∠A