In a ` A B C ,\ /_A=50^0a n d\ B C` is produced to a point `Ddot` If the bisectors of `/_A B C\ a n d\ /_A C D` meet at `E ,` then `/_E=` `25^0` (b) `50^0` (c) `100^0` (d) `75^0`

Explanation

In the given triangle, /_ACD = /_A + /_B.
(Exterior angle is equal to the sum of two opposite interior angles.) We know that the sum of all three angles of a triangle is 180°.
Therefore, for the given triangle, we know that the sum of the angles = `180°
∠ABC + ∠BCA + ∠CAB = 180°
∠A + ∠B + ∠BCA = 180°
∠BCA = 180° – (∠A + ∠B)
But we know that EC bisects ∠ACD
Therefore ∠ECA = ∠ACD/2
∠ECA = (∠A + ∠B)/2
[∠ACD = (∠A + ∠B)]` But EB bisects `∠ABC
∠EBC = ∠ABC/2 = ∠B/2
∠EBC = ∠ECA + ∠BCA
∠EBC = (∠A + ∠B)/2 + 180° – (∠A + ∠B)
If we use same steps for △EBC, then we get, ∠B/2 + (∠A + ∠B)/2 + 180° – (∠A + ∠B) + ∠BEC = 180° ∠BEC = ∠A + ∠B – (∠A + ∠B)/2 – ∠B/2 ∠BEC = ∠A/2 =25@`.