Question

prove that the angle between the bisectors of two exterior angles of a triangle is complementary of the half of the third angle.

Answer 1

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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