in triangle ABC, angle A =120°, AB=a , AC = b , if AD is angle bisector of angle A and D is point on BC. Prove that AD= ab/a+b
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1
Answer:
120⁰+AB+AC+BC+AD
= 120⁰+AB
=120⁰ab
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If <BAC = 120 ,<BAD=60,<CAD=60....area of ABC= Area of ABD+Area of ADC. ..........(1) Here AB =a, AC = b, AD = h... Then from equation (1).... 1/2absin120=1/2 ah sin60 +1/2 bh sin60. 1/2ab×√3/2 =1/2 ah ×√3/2+12bh×√3/2. (here sin120=√3/2,sin60=√3/2) 1/2×√3/2×ab =1/2×√3/2(ah+bh) ...(2) by cancellation 1/2×√3/2 from both Side of equation (2) we get. ab=ah+bh. i. e. h=ab/a+b proved explanation:
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