Question

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

Answer 1

By exterior angle theorem,

∠ACD = ∠A + ∠B

∠ACD = 68° + ∠B

\(\frac{1}{2}\)∠ACD = 34° + \(\frac{1}{2}\)∠B

34° = \(\frac{1}{2}\)∠ACD - ∠EBC (i)

Now,

In ΔBEC

∠ECD = ∠EBC + ∠E

∠E = ∠ECD - ∠EBC

∠E = \(\frac{1}{2}\)∠ACD - ∠EBC (ii)

From (i) and (ii), we get

∠E = 34°