Question

In △ABC, ∠A = 50° and BC is produced to a point D. The bisectors of ∠ABC and ∠ACD meet at E. Find ∠E.

Answer 1

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In Δ ABC, Using exterior angle theorem :

ext. ∠ACD = ∠A + ∠B

½ ext. ∠ACD = ½ ∠A + ½ ∠B

½ (2∠2 ) = ½ ∠A + ½ (2∠1)

[∵ BE & CE are bisectors of ∠B & ∠ACD, i.e ∠B = 2∠1 , ∠ACD = 2∠2]

∠2 = ½ ∠A + ∠1 ………(1)

In Δ BCE, Using exterior angle theorem :

ext. ∠ACD = ∠1 + ∠E

∠2 = ∠1 + ∠E ……….(2)

From eq 1 and 2 ,

½ ∠A + ∠1 = ∠1 + ∠E

½ ∠A = ∠E

∠E = ½ ∠A

∠E = ½ × 50°

[Given ∠A = 50°]

∠E = 25°

Hence the value of ∠E is 25° .

Among the given options option (A) 25° is correct.

Answer 2

Answer:

In Δ ABC, Using exterior angle theorem :

ext. ∠ACD = ∠A + ∠B

½ ext. ∠ACD = ½ ∠A + ½ ∠B

½ (2∠2 ) = ½ ∠A + ½ (2∠1)

[∵ BE & CE are bisectors of ∠B & ∠ACD, i.e ∠B = 2∠1 , ∠ACD = 2∠2]

∠2 = ½ ∠A + ∠1 ………(1)

In Δ BCE, Using exterior angle theorem :

ext. ∠ACD = ∠1 + ∠E

∠2 = ∠1 + ∠E ……….(2)

From eq 1 and 2 ,

½ ∠A + ∠1 = ∠1 + ∠E

½ ∠A = ∠E

∠E = ½ ∠A

∠E = ½ × 50°

[Given ∠A = 50°]

∠E = 25°

Hence the value of ∠E is 25° .

Among the given options option (A) 25° is correct.