Question

the side BC of a triangle ABC is produced to ray BC such that D is on ray BC the bisector of angle A meets BC in L prove that angle abc + angle ACD is equal to 2 angle A L C​

Answer 1

ABC +  ∠ACD = 2∠ALC

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

Answer 2

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