Question

Bisection of angles A,B and C of a triangle ABC intersect to circumfarance of D,E and F respectivly. Prove that the angle of $\Delta DEF$ are ${90}^{\circ }-\frac{1}{2}A$ and ${90}^{\circ }-\frac{1}{2}C$

Answer 1

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