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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.

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Grade:9

3 Answers

Yash Mandil
23 Points
7 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
7 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
2 years 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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