Question

bisector 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 d e f are 90 degree minus half angle A 90 degrees minus half Angle B and 90 degree minus half angle C

Answer 1

Answer:

Step-by-step explanation:

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.

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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

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 Hope this will help you....

Answer 2

Answer:

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)]