Question

ABC is a triangle with BC produced to D.Also bisectors of angle ABC and angle ACD meet at E. show that angle BEC=1/2 angle BAC

Answer 1

Answer:

Given: ABC is a triangle,

In which $D\in BC$

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.

Answer 2

∠BEC=1/2 ∠BAC

Step-by-step explanation:

Given,

ABC is a Δ,

with BE and CE being the angle bisectors of ∠ABC and ∠ACD,

To prove,

∠BEC = 1/2 ∠BAC

As we know,

BE being the angle bisector of ∠ ABC,

∵ ∠ABE = ∠EBC

&, CE being the angle bisector of ∠ACD,

∵ ∠ACE = ∠ECD

By employing the exterior angle theorem,

∠BAC  + ∠ABC  

= ∠ACD

∵ (∠ECD + ∠ACE) = (∠EBC + ∠ABE) + ∠BAC

⇒ 2 * ∠ECD =  ∠BAC  + 2 *∠EBC

⇒ ∠BAC = 2(∠ECD - ∠EBC) -(1)

by using the exterior angle theorem again,

∠ECD = ∠BEC + ∠EBC

= ∠BEC = ∠ECD - ∠EBC -(2)

Through using equations (1) and (2),

∵  ∠BAC = 2 ∠BEC

⇒ ∠BEC  = 1/2 ∠BAC

Learn more: Angle Bisectors

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