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bisectors of angle A B and C of a triangle ABC intersect its circumcircle at D , E and F respectively . Prove that the angles of the triangle are 90 -1/2 A , 90 – ½ B and 90 -1/2 C.

bisectors of angle A B and C of a triangle ABC intersect its circumcircle at D , E and F respectively . Prove that the angles of  the triangle are 90 -1/2 A , 90 – ½ B and 90 -1/2 C.

Question Image
Grade:9

3 Answers

Yash Mandil
23 Points
6 years ago
It is given that BE is the bisector of ∠B
.: ∠ABE = ∠B/2   
However, ∠ADE = ∠ABE (Angles in the same segment for chord AE)
  • ∠ADE = ∠B/2
Similarly, ∠ACF = ∠ADF =  ∠C/2 (Angle in the same segment for chord AF) 
 ∠D = ∠ADE + ∠ADF
 ∠D = ∠B/2  + ∠C/2
      = ½  ( ∠B + ∠C )
    = ½  (  180 – ∠A )        {since ∠A + ∠B + ∠C =180 ; angle sum property }
    = 90 – ∠A/2 
Similarly, it can be proved that
∠E = 90 – ∠B/2     &
∠F = 90 – ∠C/2
 
 
Yash Mandil
23 Points
6 years ago
 this answer was asked by another student but i can not insert image that’s why i asked same question with image  
Rishi Sharma
askIITians Faculty 646 Points
one year ago
Dear Student,
Please find below the solution to your problem.

It is given that BE is the bisector of ∠B
.: ∠ABE = ∠B/2
However, ∠ADE = ∠ABE (Angles in the same segment for chord AE)
∠ADE = ∠B/2
Similarly, ∠ACF = ∠ADF = ∠C/2 (Angle in the same segment for chord AF)
∠D = ∠ADE + ∠ADF
∠D = ∠B/2 + ∠C/2
= ½ ( ∠B + ∠C )
= ½ ( 180 – ∠A ) {since ∠A + ∠B + ∠C =180 ; angle sum property }
= 90 – ∠A/2
Similarly, it can be proved that
∠E = 90 – ∠B/2 &
∠F = 90 – ∠C/2

Thanks and Regards


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