Question

tell the correct answer

Answer 1

the side of BC of triangle ABC is produced to D If the bisector of angle A meets BC in E. prove that angle abc + angle acd = 2 angle AEC

Solution:

the bisector of angle A meets BC in E

=> ∠BAE = ∠CAE = x

Let say ∠ABC = ∠ABE = α  ( as E is on BC)

∠AEC = ∠ABE + ∠BAE

=> ∠AEC = α + x

∠ACD = ∠AEC + ∠CAE

=> ∠ACD = α + x + x

∠ABC +  ∠ACD  = α + α + x + x

=> ∠ABC +  ∠ACD = 2 ( α + x)

=> ∠ABC +  ∠ACD = 2∠AEC