Prove that the angle between internal bisector of one base angle and the external bisector
of the other base angle of a triangle is equal to one half of the vertical angle.
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Step-by-step explanation:
In △ABC, let ∠ABD be the exterior angle.
Let ray BE be bisector of interior ∠ABC and ray BF be bisector of exterior ∠ABD.
∴∠ABE=∠EBC and ∠ABF=∠FBD let ∠ABE=∠EBC=x ...(1)
let ∠ABF=∠FBD=x ...(2)
We know, ∠ABC+∠ABD=180
o
....angles in linear pair
∴(∠ABE+∠EBC)+(∠ABF+∠FBD)=180
o
....angle addition property
∴(x+x)+(y+y)=180
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...from (1) & (2)
∴2x+2y=180
o
Dividing by 2 on both sides we get,
x+y=90
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∴∠ABE+∠ABF=90
o
....from (1) & (2)
solution
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