Ex 10.6, 8 (Optional)
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.
Circumcircle is a circle where all 3 vertices of triangle are on the circle
Given: AD, BE and CF are bisector of angles A, B and C of ∆ABC
To Prove: Angles of ∆DEF are 90° − 1/2 ∠ A, 90° − 1/2 ∠B and 90° − 1/2 ∠C
Proof:
Given that
AD is bisector of ∠A
Let ∠ 1 = (∠𝐴)/2
And,
BE is bisector of ∠B
Let ∠ 2 = (∠𝐵)/2
And,
CF is bisector of ∠C
Let ∠ 3 = (∠𝐶)/2
Now,
We know that
Angles in the same segment are equal
Segment BD,
suspends ∠BAD and ∠BED on the circle.
∴ They be equal.
i.e. ∠BAD = ∠BED = ∠ 1
Similarly
For segment CD,
∠CAD = ∠CFD = ∠1
For segment AE,
∠ABE = ∠ADE = ∠2
For segment CE,
∠CBE = ∠CFE = ∠2
For segment AF
∠ACF = ∠ADF = ∠3
For segment FB
∠FCB = ∠FEB = ∠3
So, our figure looks like
Now,
In Δ ABC,
By angle sum property
∠A + ∠B + ∠C = 180°
2∠1 + 2∠2 + 2∠3 = 180°
2 (∠1 + ∠2 + ∠3) = 180°
∠1 + ∠2 + ∠3 = (180°)/2
∠1 + ∠2 + ∠3 = 90°
Now,
In Δ DEF,
By angle sum property
∠D + ∠E + ∠F = 180°
∠D + ∠3 + ∠1 + ∠2 + ∠1 = 180°
∠D + ∠1 + (∠1 + ∠2 + ∠3) = 180°
∠D + ∠1 + 90° = 180°
∠D = 180° − 90° − ∠1
∠D = 90° − ∠1
Putting ∠ 1 = (∠𝐴)/2
∠D = 90° − (∠𝐴)/2
Similarly, we can prove
∠E = 90° − (∠𝐵)/2
∠F = 90° − (∠𝐶)/2
Hence Proved

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 13 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.

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