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

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Solution

$In the given triangle, ∠ ACD = ∠ A + ∠ B . (Exterior angle is equal to the sum of two opposite interior angles .) We know that the sum of all three angles of a triangle is 180 ° . Therefore, for the given triangle, we can say that : ∠ ABC + ∠ BCA + ∠ CAB = 180 ° . (Sum of all angles of △ ABC) ∠ A + ∠ B + ∠ BCA = 180 ° ∠ BCA = 180 ° - (∠ A + ∠ B) ∠ ECA = \frac{∠ ACD}{2} (∵ EC bisects ∠ ACD) ∠ ECA = \frac{∠ A + ∠ B}{2} (∵ ∠ ACD = ∠ A + ∠ B) ∠ EBC = \frac{∠ ABC}{2} = \frac{∠ B}{2} (∵ EB bisects ∠ ABC) ∠ ECB = ∠ ECA + ∠ BCA \Rightarrow ∠ ECB = \frac{∠ A + ∠ B}{2} + 180 ° - (∠ A + ∠ B) If we use the same logic for △ EBC, we can say that : ∠ EBC + ∠ ECB + ∠ BEC = 180 ° (Sum of all angles of △ EBC) \frac{∠ B}{2} + \frac{∠ A + ∠ B}{2} + 180 ° - (∠ A + ∠ B) + ∠ BEC = 180 ° ∠ BEC = ∠ A + ∠ B - (\frac{∠ A + ∠ B}{2}) - \frac{∠ B}{2} ∠ BEC = \frac{∠ A}{2} \Rightarrow ∠ BEC = \frac{50 °}{2} = 25 °$

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