In Δ ABC, ∠A=50° and BC is produced to a point D. If the bisectors of ∠ABC and ∠ACD meet at E, then ∠E =
A. 25°
B. 50°
C. 100°
D. 75°
Answers
Given: In Δ ABC, ∠A = 50° and BC is produced to a point D. If the bisectors of ∠ABC and ∠ACD meet at E.
To Find : ∠E
Proof :
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.
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Similar questions :
In a Δ ABC, If ∠A = 60°, ∠B =80° and the bisectors of ∠B and ∠C meet at O, then ∠BOC= A. 60° B. 120° C. 150° D. 30°
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Side BC of a triangle ABC has been produced to a point D such that ∠ACD = 120°. If ∠B = 1/2 ∠A, then ∠A is equal to
A. 80°
B. 75°
C. 60°
D. 90°
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Answer:
Step-by-step explanation:
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°