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shubhamsrg

  • 2 years ago

ABC is a triangle where CD is the angle bisector . prove that geometric mean of AC and BC is greater than CD

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  1. shubhamsrg
    • 2 years ago
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    |dw:1350477585294:dw|

  2. estudier
    • 2 years ago
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    Cosine Law maybe...?

  3. shubhamsrg
    • 2 years ago
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    i didnt give this question more than 10 mins,,thought i'd discuss here directly..so i dont know if that may work,, but i am thinking of some use of angle bisector theorem..

  4. estudier
    • 2 years ago
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    |dw:1350477770138:dw|

  5. estudier
    • 2 years ago
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    bc = a^2-b^2-c^2/Cos A greater than CD^2 (from GM > CD)

  6. estudier
    • 2 years ago
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    That's on whole triangle, could maybe get more from the 2 smaller triangles..

  7. shubhamsrg
    • 2 years ago
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    am so very sorry @estudier i had mistyped the question then,,please see now..

  8. estudier
    • 2 years ago
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    OK, so ab instead of bc, same thing though...

  9. shubhamsrg
    • 2 years ago
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    okay so we have c^2 = a^2 + b^2 - 2ab cosC => ab = a^2 + b^2 - c^2 / 2cos C

  10. shubhamsrg
    • 2 years ago
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    the next step ?

  11. estudier
    • 2 years ago
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    I haven't looked, I am just speculating (because of the ab term). Maybe you can apply the cosine law to the 2 smaller triangles and using the bisected angle

  12. estudier
    • 2 years ago
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    |dw:1350478392349:dw|

  13. shubhamsrg
    • 2 years ago
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    |dw:1350478324826:dw| x^2 = b^2 + m^2 -2bm cos@ c^2 + x^2 - 2cx = m^2 + a^2 - 2am cos@ =>c^2 - 2cx = a^2 -b^2 - 2mcos@ (a+b)

  14. shubhamsrg
    • 2 years ago
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    we are missing the ab term..hmm..

  15. estudier
    • 2 years ago
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    Doesn't matter, the ab term is in the first equation....

  16. shubhamsrg
    • 2 years ago
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    and we had c^2 = a^2 + b^2 - 2ab cos2@ how shall we combine ?

  17. estudier
    • 2 years ago
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    Well, the other half is m^2 so may as well isolate that ...

  18. estudier
    • 2 years ago
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    Then compare the 2 sides, see how to reduce them...

  19. shubhamsrg
    • 2 years ago
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    if we subtract , we have 2cx = 2b^2 - 2 ( ab cos2@ + m(a+b)cos@ )

  20. shubhamsrg
    • 2 years ago
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    why so ?

  21. sara12345
    • 2 years ago
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    |dw:1350478727702:dw|

  22. shubhamsrg
    • 2 years ago
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    i'd second @sara12345

  23. estudier
    • 2 years ago
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    Someone lay me out ab > m^2 -> Using the expressions we have?

  24. estudier
    • 2 years ago
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    I was using my letters, now they are all different.....

  25. estudier
    • 2 years ago
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    @shubhamsrg Use your diagram, your letters......

  26. sauravshakya
    • 2 years ago
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    |dw:1350479111593:dw|

  27. sauravshakya
    • 2 years ago
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    AC cosy = BC cos(2x-y) = CD cos (x-y)

  28. sauravshakya
    • 2 years ago
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    This gives:|dw:1350479343315:dw|

  29. shubhamsrg
    • 2 years ago
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    why AC cosy = BC cos(2x-y) = CD cos (x-y) ??

  30. sauravshakya
    • 2 years ago
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    Express AC, BC and CD interms of CE

  31. shubhamsrg
    • 2 years ago
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    you interchanged AC and BC by mistake there it seems..please confirm..

  32. ganeshie8
    • 2 years ago
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    |dw:1350505825096:dw|

  33. ganeshie8
    • 2 years ago
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    \((bk)^2 = b^2+CD^2 - 2.b.CD \cos C/2\) \((ak)^2 = a^2+CD^2 - 2.a.CD \cos C/2\) combining above two : \(a[b^2(k^2-1) - CD^2] = b[a^2(k^2-1)-CD^2]\) \(CD^2(b-a) = ab(k^2-1)(a-b)\) \(CD^2 = ab(1-k^2)\) \(CD < \sqrt{ab}\)

  34. sauravshakya
    • 2 years ago
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    1 Attachment
  35. ganeshie8
    • 2 years ago
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    this proof looks more elegant, it didnt even depend on angle-bisector theorem !

  36. sauravshakya
    • 2 years ago
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    Is it correct?

  37. ganeshie8
    • 2 years ago
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    looks flawless to me :)

  38. sara12345
    • 2 years ago
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    it does we cancelled sinx in the first line becoz the angles are same

  39. sauravshakya
    • 2 years ago
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    Oh ya..... it does.

  40. shubhamsrg
    • 2 years ago
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    thanks a ton again @sauravshakya and @ganeshie8 earlier i couldnt find 1 method but now i have 2 ^_^

  41. sauravshakya
    • 2 years ago
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    Welcome buddy.

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