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° – (½)A, 90° – (½)B and 90° – (½)C.
Refer to the following diagram :

Answer 1

Answer:

The region between a chord and either of its arcs is called a segment the circle.

Angles in the same segment of a circle are equal.

Step-by-step explanation:

Let bisectors of angles A, B and C of a triangle ABC intersect its circumcircle at D, E and F . 

Now from figure,

∠D = ∠EDF

∠D = ∠EDA + ∠ADF

Since ∠EDA and ∠EBA are the angles in the same segment of the circle.

So  ∠EDA  = ∠EBA

Hence  ∠D = ∠EBA + ∠FCA

Again  ∠ADF and ∠FCA are the angles in the same segment of the circle. 

hence ∠ADF = ∠FCA

Again since BE is the internal bisector of ∠B and CF is the internal bisector of ∠C

So ∠D = ∠B/2 + ∠C/2

Similarly

∠E = ∠C/2 + ∠A/2

and 

∠F = ∠A/2 + ∠B/2

Now ∠D = ∠B/2 + ∠C/2

=∠D = (180 - ∠A)/2

 (∠A + ∠B + ∠C = 180)

∠D = 90 - ∠A/2

∠E = (180 - ∠B)/2

∠E = 90 - ∠B/2

and ∠F = (180 - ∠C)/2

∠F = 90 - ∠C/2