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 = 90o - 1/2 angle A.
As we know in the given problem BD subtends angles ∠BED and ∠BAD
AF subtends angles ∠ADF and ∠ACF∠ ADF=∠ACF=∠C/2
=1/2(180-∠A) ( since ∠A+∠B+∠C=180º )
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