Question

Bisector of angles A,B and C of a triangleABC intersect its circum circle at D.,E and F.Prove that the angles of the triangle DEF are 90-A/2,90-B/2,90-C/2

Answer 1

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

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 ∠ECA are the angles in the same segment of the circle. 

hence ∠ADF = ∠ECA

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

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Answer 2

Hey mate!!!!
Here's ur answer:


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

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 ∠ECA are the angles in the same segment of the circle. 

hence ∠ADF = ∠ECA

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

Hope u understand
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