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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 =
Options
25°
50°
100°
75°
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Solution
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°`
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