The bisector of angle A meets at BC in L in triangle ABC.
prove that <ABC + <ACB = 2<ALC
plss solve it ASAP
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1
SIDE BC OF TRIANGLE PRODUCED TO D.
LET <BAL AND <LBA BE a AND b RESPECTIVELY.
BY EXTERIOR ANGLE PROPERTY,
<ALC=a+b
IN TRIANGLE ABC
<ACD=2a+b
<ABC+<ACD
=b+2a+b
=2(<ALC)
LET <BAL AND <LBA BE a AND b RESPECTIVELY.
BY EXTERIOR ANGLE PROPERTY,
<ALC=a+b
IN TRIANGLE ABC
<ACD=2a+b
<ABC+<ACD
=b+2a+b
=2(<ALC)
Answered by
2
Done I think you have done a mistake in question
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