In Fig., side BC of Δ ABC is produced to point D such that bisectors

1.	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» 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°

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»

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°