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प्रश्न
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 Δ DEF are 90° - `"A"/2` , 90° - `"B"/2` and 90° - `"C"/2` respectively.
उत्तर
Since AD , BE and CF are bisectors of ∠ A , ∠ B and ∠ C respectively.
∴ ∠1 = ∠ 2 = ∠`"A"/2`
∠3 = ∠4 = ∠`"B"/2`
∠5= ∠6 = ∠`"C"/2`
∠ADE = ∠3 ....(1)
Also ∠ADF = ∠6 ....(2) (angles in the same segment)
Adding (1) and (2)
∠ADE + ∠ADF = ∠3 + ∠6
∠D = `1/2`∠B + `1/2` ∠C
∠D = `1/2` (B + ∠C) = `1/2` (180 - ∠A) (∠A + ∠B + ∠C = 180°)
∠D = 90 - `1/2` ∠A
Similarly ,
∠E = 90 - `1/2` ∠B , ∠F = 90 - `1/2` ∠C
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