Question

The side BC of triangle ABC is produced to D. The bisector of angle A meets BC on E. prove that angle ABC +angle ACD =2 angle AEC

Answer 1

Step-by-step explanation:

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

the bisector of angle A meets BC in L

=> ∠BAL  = ∠CAL = x

Let say ∠ABC = ∠ABL = α  ( as L is on BC)

∠ALC = ∠ABL + ∠BAL

=> ∠ALC = α + x

∠ACD = ∠ALC + ∠CAL

=> ∠ACD = α + x + x

∠ABC +  ∠ACD  = α + α + x + x

=> ∠ABC +  ∠ACD = 2 ( α + x)

=> ∠ABC +  ∠ACD = 2∠ALC