in triangle ABC if the angle bisectors of base angle ABC and of exterior angle ACD meet at point E. if the angle BAC =80° then angle BEC will be
please answer with steps I will mark you as brainliest
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Answer:
it will be half of angle a
answer is 40°
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Step-by-step explanation:
Given: ABC is a triangle,
In which D\in BCD∈BC
Also, BE and CE are the angle bisector of the angles ABC and ACD,
We have to prove that: ∠BEC = 1/2 ∠BAC
Proof:
BE is the angle bisector of angle ABC,
⇒ ∠ABE = ∠EBC
And, CE is the angle bisector of angle ACD,
⇒ ∠ACE = ∠ECD
By the exterior angle theorem,
∠ACD = ∠ABC + ∠BAC
⇒ (∠ACE + ∠ECD) = (∠ABE + ∠EBC) + ∠BAC
⇒ 2∠ECD = 2∠EBC + ∠BAC
⇒ ∠BAC = 2(∠ECD - ∠EBC) ---------(1)
Now, again by exterior angle theorem,
∠ECD = ∠EBC+∠BEC
⇒ ∠BEC = ∠ECD - ∠EBC ------------(2)
By equation (1) and (2),
∠BAC = 2 ∠BEC
⇒ 1/2 ∠BAC = ∠BEC
Hence, proved.
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