Question

8. In fig side BC of AABC is produced to point D such that bisectors of < ABC and <
ACD meet at a point E. If < BAC = 68°, find <BEC.​

Answer 1

Answer:

By exterior angle theorem,

∠ACD = ∠A + ∠B

∠ACD = 68° + ∠B

1/2∠ACD = 34° + 1/2 ∠B

34°= 1/2∠ACD - 1/2∠EBC (i)

Now,In ∆BEC

∠ECD = ∠EBC + ∠E

∠E=∠ECD+∠EBC

∠E = 1/2∠ACD - ∠EBC (ii)

From (i) and (ii), we get

∠E = 34°

${} {\huge {\mathfrak{\red {\underline{hope \: it \: helps \: you}}}}}$