Question

the side AB of triangle ABC is produced to point D. if the bisectors of angle CAB and angle CBD meets at E, then prove that angle AEB=1/2angle ACB.

Answer 1

hope it will help u....

Answer 2

∠AEB = ∠ACB/2 the side AB of triangle ABC is produced to point D. if the bisectors of angle CAB and angle CBD meets at E

Step-by-step explanation:

Exterior angle of triangle = Sum of two opposite internal angles

=> ∠3 + ∠4  = ∠1 + ∠2 + ∠6

∠3 = ∠4  

∠1 = ∠2

=>  ∠4 + ∠4  = ∠2 + ∠2 + ∠6

=> 2 ∠4 = 2 ∠2  + ∠6

∠4 = ∠2 + ∠5

=> 2 (∠2 + ∠5) =  2 ∠2  + ∠6

=> 2∠2 + 2∠5  =  2 ∠2  + ∠6

=>  2∠5  =  ∠6

=> ∠5  =  ∠6 /2

∠5 = ∠AEB

∠6 = ∠ACB

=> ∠AEB = ∠ACB/2

QED

Proved

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