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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 = 90° - `1/2` `∠`A - ICSE Class 10 - Mathematics

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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  = 90° - `1/2` `∠`A

Solution

Join ED, EF and DF. Also join BF, FA, AE and EC.
`∠`EBF =  `∠`ECF =  `∠`EDF …………(i) (angles in the same segment)
In cyclic quadrilateral AFBE,
`∠`EBF + `∠`EAF =180° ……………(ii) (sum of opposite angles)
Similarly in cyclic quadrilateral CEAF,
`∠`EAF + `∠`ECF =180° …………. (iii)
Adding (ii) and (iii)
⇒ `∠`EDF + `∠`ECF + 2`∠`EAF = 360°
⇒ `∠`EDF +` ∠`EDF + 2`∠`EAF =360°    (from (i))

 ⇒ 2 `∠`EDF  + 2`∠` EAF =  360°

⇒ `∠`EDF + `∠`EAF  = 180°
⇒  `∠`EDF +  `∠`1  + `∠`BAC + `∠`2 = 180°
But `∠`1  = `∠`3 and `∠`2 = `∠`4
(angles in the same segment)

∴ `∠EDF+ `∠`3 + `∠`BAC + `∠`4  = 180°`
But  `∠`4 =`1/2` ∠C, ∠3 = `1/2 ` ∠B
∴ `∠EDF + 1/2  ∠B  + ∠BAC + 1/2  ∠C  = 180°`

⇒ `∠ EDF + 1/2  ∠B  + 2xx1/2 ∠A + 1/2 ∠C  = 180°`
⇒ ∠`EDF + 1/2 ( ∠A  + ∠B  + ∠C) + 1/2 ∠A = 180`
⇒ ∠`EDF + 1/2 (180°) + 1/2  ∠A  = 180°`
⇒ ∠`EDF + 90 + 1/2  ∠A =  180°`
⇒ ∠`EDF =  180° - ( 90 + 1/2 ∠ A)`
⇒ ∠`EDF = 180° - 90° 1/2 ∠A`

⇒∠`EDF = 90°- 1/2 ∠A`

 

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Solution 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 = 90° - `1/2` `∠`A Concept: Cyclic Properties.
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