Prove that the angle between the bisectors of two exterior angles of a triangle is complementary of the half of the third angle.Pls Answer!
Answers
Given : a triangle & angle between bisector of exterior angles
To find : prove that the angle between the bisectors of two exterior angles of a triangle is complementary of the half of the third angle.
Solution:
Let say triangle ABC
Consider bisector angle of B & C meeting at D & forming Δ BCD
∠B' = Exterior angle of B
∠C' = Exterior angle of C
∠B' = ∠C + ∠A
∠C' = ∠B + ∠A
∠CBD = ∠B' /2 = (∠C + ∠A)/2
∠BCD = ∠C' /2 = (∠B + ∠A)/2
in Δ BCD
∠CBD + ∠BCD + ∠CDB = 180°
∠CDB = ( angle between the bisectors of two exterior angles of a triangle )
(∠C + ∠A)/2 + ∠A /2 + ∠CDB = 180°
=> (∠C + ∠A +∠B )/2 + ∠A /2 + ∠CDB = 180°
=> (180° )/2 + ∠A /2 + ∠CDB = 180°
=> 90° + ∠A /2 + ∠CDB = 180°
=> ∠A /2 + ∠CDB = 90°
=> ∠A /2 & ∠CDB are complementary angles
QED
Hence proved
angle between the bisectors of two exterior angles of a triangle is complementary of the half of the third angle.
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