Bisectors of angles A, B and C of a triangle ABC intersect its circumcircle at D, E and F respectively.
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20/07/2021 4:53 pm
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Bisectors 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 DEF are 90°–(1/2)A, 90°–(1/2)B and 90°–(1/2)C.
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20/07/2021 4:55 pm
Consider the following diagram
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)
By adding equations (i) and (ii) we get,
∠FDA+∠EDA = ∠FCA+∠EBA
Or, ∠FDE = ∠FCA+∠EBA = (1/2)∠C + (1/2)∠B
We know, ∠A + ∠B + ∠C = 180°
So, ∠FDE = (1/2)[∠C+∠B] = (1/2)[180°-∠A]
∠FDE = [90-(∠A/2)]
In a similar way,
∠FED = [90° -(∠B/2)]
∠EFD = [90° -(∠C/2)]
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