Question

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.​

Answer 1

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

o

...from (1) & (2)

∴2x+2y=180

o

Dividing by 2 on both sides we get,

x+y=90

o

∴∠ABE+∠ABF=90

o

....from (1) & (2)

solution