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  Δ DEF are 90° - ${\displaystyle \frac{\text{A}}{2}}$ , 90° - ${\displaystyle \frac{\text{B}}{2}}$ and 90° - ${\displaystyle \frac{\text{C}}{2}}$ respectively.

Answer 1

Since AD , BE and CF are bisectors of ∠ A , ∠ B and ∠ C respectively.

∴ ∠1 = ∠ 2 = ∠${\displaystyle \frac{\text{A}}{2}}$

∠3 = ∠4 = ∠${\displaystyle \frac{\text{B}}{2}}$

∠5= ∠6 = ∠${\displaystyle \frac{\text{C}}{2}}$

∠ADE = ∠3  ....(1)

Also ∠ADF = ∠6  ....(2)  (angles in the same segment)

Adding (1) and (2)

∠ADE + ∠ADF = ∠3 + ∠6

∠D = ${\displaystyle \frac{1}{2}}$∠B + ${\displaystyle \frac{1}{2}}$ ∠C

∠D = ${\displaystyle \frac{1}{2}}$ (B + ∠C) = ${\displaystyle \frac{1}{2}}$ (180 - ∠A) (∠A + ∠B + ∠C = 180°)

∠D = 90 - ${\displaystyle \frac{1}{2}}$ ∠A

Similarly ,

∠E = 90 - ${\displaystyle \frac{1}{2}}$ ∠B , ∠F = 90 - ${\displaystyle \frac{1}{2}}$ ∠C