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Questions
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°-A, 90° − `1/2 A, 90° − 1/2 B, 90° − 1/2` C.
Bisectors of angles A, B and C of a triangle ABC intersect its circumcircle at points D, E and F respectively. Prove that angles of ΔDEF are 90° − `1/2 ∠A, 90° − 1/2 ∠B, 90° − 1/2` ∠C.
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Solution
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
`= (angleB)/2 + (angleC)/2`
`= 1/2(angleB + angleC)`
`= 1/2(180^@ - angleA)`
`= 90^@ - 1/2angleA`
Similarly, it can be proved that
`angleE = 90^@ - 1/2angleB`
`angleF = 90^@ - 1/2angleC`
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