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, 90° - 1/2 C
Solution:
We know that angles in the same segment are equal.
A diagram is constructed as per the given question.
It is given that BE is the bisector of ∠B, AD is the bisector of ∠A and CF is the bisector of ∠C.
Thus, ∠ABE = ∠B/2
However, ∠ADE = ∠ABE (Angles in the same segment for chord AE)
Thus, ∠ADE = ∠B/2
Similarly, ∠ADF = ∠ACF = ∠C/2 (Angle in the same segment for chord AF)
∠D = ∠ADE + ∠ADF
= ∠B/2 + ∠C/2 [Since ∠ADE = ∠B/2 and ∠ADF = ∠C/2]
= 1/2 (∠B + ∠C )
= 1/2 (180° - ∠A) [Angle sum property of triangle ABC]
= 90° - 1/2 A
Similarly, it can be proved for
∠E = 90° - 1/2 B
∠F = 90° - 1/2 C.
Thus we have proved that the angles of the triangle DEF are 90° - 1/2 A, 90° - 1/2 B, 90° - 1/2 C.
☛ Check: NCERT Solutions for Class 9 Maths Chapter 10
Video Solution:
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, 90° - 1/2 C
Maths NCERT Solutions Class 9 Chapter 10 Exercise 10.6 Question 8
Summary:
If the bisectors of angles A, B and C of a triangle ABC intersect its circumcircle at D, E and F respectively, we have proved that the angles of the triangle DEF are 90° - 1/2 A, 90° - 1/2 B, 90° - 1/2 C
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