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Question
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^@-1/2A, 90^@-1/2B" and "90^@-1/2C`
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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