Question

prove that the bisectors of the exterior angles of the base of a triangle enclose an angle equal to 90 + half the vertical angle

Answer 1

To prove- ∠E=12∠A

By exterior angle theorem

∠A+∠B=∠ACD

12∠A+12∠B=12∠ACD

12∠A+∠1=∠2

∠2=∠1+12∠A−(1)

In△BCE

∠ECD=∠1+∠E

∠2=∠1+∠E−(2)

From equation 1 and 2

∠1+∠E=∠1+12∠A

∠E=12∠A.

Answer 2

Answer:

$\large\bf\underline \blue{To \: Prove-}$- ∠E=12∠A

By exterior angle theorem

∠A+∠B=∠ACD

12∠A+12∠B=12∠ACD

12∠A+∠1=∠2

∠2=∠1+12∠A−(1)

In△BCE

∠ECD=∠1+∠E

∠2=∠1+∠E−(2)

$\large\bf\underline \blue{-From \: equation \: 1 \: and \: 2}</p><p>$

∠1+∠E=∠1+12∠A

∠E=12∠A.