Question

in the given figure ,the side BC of triangle ABC has been produced to a point D .If the bisectors of ANGLE ABC and ANGLE ACD meet at point E then prove that ANGLE BEC
=1/2 angle BAC​

Answer 1

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Answer 2

Answer:

∠BEC = 1/2∠BAC is proved.

Step-by-step explanation:

Given:-

BE is the bisector of ∠ABC, i.e., ∠ABE = ∠EBC.

Also, CE is the bisector of ∠ACD, i.e., ∠ACE = ∠ECD.

To prove:-

∠BEC = 1/2∠BAC

From the figure,

In ΔBEC,

∠ECD = ∠BEC + ∠EBC (Exterior angle property)

∠BEC = ∠ECD - ∠EBC . . . . . (1)

Similarly,

In ΔABC,

∠ACD = ∠ABC + ∠BAC (Exterior angle property)

∠BAC = ∠ACD - ∠ABC . . . . . (2)

Since CE is the bisector of ∠ACD.

⇒ ∠ACD = ∠ACE + ∠ECD

⇒ ∠ACD = ∠ECD + ∠ECD (As ∠ACE = ∠ECD)

⇒ ∠ACD = 2∠ECD

Also,

BE is the bisector of ∠ABC.

⇒ ∠ABC = ∠ABE + ∠EBC

⇒ ∠ABC = ∠EBC + ∠EBC (As ∠ABE = ∠EBC)

⇒ ∠ABC = 2∠EBC

Substitute the values of ∠ACD and ∠ABC in the equation (2), we get

∠BAC = 2∠ECD - 2∠EBC

∠BAC = 2(∠ECD - ∠EBC) . . . . . (3)

From (1) and (3), we have

∠BAC = 2∠BEC

1/2∠BAC = ∠BEC

Hence proved.

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