Question

In the given figure, side BC of ΔABC is produced to point D such that bisectors of ∠ABC and ∠ACD meet at a point E. If ∠BAC = 68°, find ∠BEC.

Answer 1

In the given figure, bisectors of ∠ABC and ∠ACD meet at E and  ∠BAC = 68° 

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.

In ΔABC with  ∠ACD as its exterior angle

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

Similarly, in ΔBE with ∠ECD as its exterior angle

exterior ∠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(68°)`

 ∠BEC = 34°

Thus,  ∠BEC = 34°