Question

In a ΔABC, ∠A = 50° and BC is produced to a point D. If the bisectors of ∠ABC and ∠ACDmeet at E, then ∠E =

विकल्प

• 25°

• 50°

• 100°

• 75°

Answer 1

In the given figure, bisectors of ∠ABCand ∠ACDmeet at E and ∠BAC = 50°

We need to find  ∠BEC

Here, using the property, “an exterior angle of the triangle is equal to the sum of the opposite interior angles”, we get,

In ΔABC with  ∠ACD as its exterior angle 

ext . ∠ACD = ∠A + ∠ABC           ........(1)

Similarly, in Δ BECwith  ∠ECDas its exterior angle

ext . ∠ECD = ∠EBC + ∠BEC  

`1/2 ∠ACD = 1/2 ∠ABC + BEC`

 (CE and BE are the bisectors of  ∠ACD and ∠ABC)

 ∠BEC = 1/2 ∠ACD - 1/2 ∠ABC     .......(2)

Now, multiplying both sides of (1) by 1/2

We get, 

`1/2 ∠ACD = 1/2 ∠A +1/2 ∠ABC`

`1/2 ∠A = 1/2 ∠ACD - 1/2 ∠ABC`    ......(3)

From (2) and (3) we get,

`∠BEC =1/2 ∠A`

`∠BEC =1/2 (50°)`

`∠BEC = 25°`

Thus, `∠BEC = 25°`