Prove the following question :

Bisectors of vertex angles A, B and C of a triangle ABC intersect its circumcircle at the points D, E and F respectively. Prove that angle EDF = 90​o - 1/2 angle A.

Dear Student,

As we know in the given problem BD subtends angles ∠BED and ∠BAD

∠BED=∠BAD=∠A/2

AF subtends angles ∠ADF and ∠ACF

∠ ADF=∠ACF=∠C/2

∠ADE=∠ABE=∠B/2

Therefore, ∠ADE+∠ADF=∠C/2+∠B/2

⇒∠EDF=1/2(∠C+∠B)

 =1/2(180-∠A)                 ( since ∠A+∠B+∠C=180º )

⇒∠EDF=90°–∠A/2



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