Question

Que: 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.​

Answer 1

Answer:

Solution:

Here, ABC is inscribed in a circle with center O and the bisectors of ∠A, ∠B and ∠C intersect the circumcircle at D, E and F respectively.

Now, join DE, EF and FD

As angles in the same segment are equal, so,

∠FDA = ∠FCA ————-(i)

∠FDA = ∠EBA ————-(i)

Adding equations (i) and (ii) we have,

∠FDA + ∠EDA = ∠FCA + ∠EBA

Or, ∠FDE = ∠FCA + ∠EBA = (½)∠C + (½)∠B

We know, ∠A + ∠B + ∠C = 180°

So, ∠FDE = (½)[∠C + ∠B] = (½)[180° – ∠A]

⇒ ∠FDE = [90 – (∠A/2)]

In a similar way,

∠FED = [90 – (∠B/2)]

And,

∠EFD = [90 – (∠C/2)]

_____________________

Hope it will be helpful :)

Answer 2

Explanation:

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