In Δ ABC , ∠ A =50° and BC is produced to a point D . If the bisectors of ∠ ABC and ∠ ACD meet at E , then ∠ E =

Question

In &#x394; ABC , &#x2220; A =50&#xB0; and BC is produced to a point D . If the bisectors of &#x2220; ABC and &#x2220; ACD meet at E , then &#x2220; E =

Philoid · Accepted Answer

In Δ ABC

∠A + ∠B + ∠C = 180^o

50^o + ∠B + ∠C = 180^o

∠B + ∠C = 180^o – 50^o

∠B + ∠C = 10^o (i)

In

∠E + ∠BCE + ∠EBC = 180^o

∠E + 180^o – ( ∠ACD) + ∠B = 180^o (ii)

By exterior angle theorem,

∠ACD = 50^o + ∠B

Putting value of ∠ACD in (ii), we get

∠E + 180^o - (50^o + ∠B) + ∠B = 180^o

∠E – 25^o - ∠B + ∠B = 0

∠E – 25^o = 0

∠E = 25^o

In Δ ABC, ∠A=50° and BC is produced to a point D. If the bisectors of ∠ABC and ∠ACD meet at E, then ∠E =