Question

pls i beg to u guys pls atleast once pls write the correct answer for this i beg u plssss i am cying :(

Answer 1

Answer:

InΔABL

let∠BAL=a&∠BA=b

exterior∠ALC=∠BAL+∠LBA

=a+b

InΔABC

ALbisect∠BAC&∠AC=a

exterior∠ACD=∠BAC+∠ABC

=2a+b

=∠ABC+∠ACo

=b+(2a+b)

=2a+2b

=2(a+b)

=2∠ALC

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Answer 2

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