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

Mathematics
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|dw:1350477585294:dw|
Cosine Law maybe...?
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..

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|dw:1350477770138:dw|
bc = a^2-b^2-c^2/Cos A greater than CD^2 (from GM > CD)
That's on whole triangle, could maybe get more from the 2 smaller triangles..
am so very sorry @estudier i had mistyped the question then,,please see now..
OK, so ab instead of bc, same thing though...
okay so we have c^2 = a^2 + b^2 - 2ab cosC => ab = a^2 + b^2 - c^2 / 2cos C
the next step ?
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
|dw:1350478392349:dw|
|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)
we are missing the ab term..hmm..
Doesn't matter, the ab term is in the first equation....
and we had c^2 = a^2 + b^2 - 2ab cos2@ how shall we combine ?
Well, the other half is m^2 so may as well isolate that ...
Then compare the 2 sides, see how to reduce them...
if we subtract , we have 2cx = 2b^2 - 2 ( ab cos2@ + m(a+b)cos@ )
why so ?
|dw:1350478727702:dw|
i'd second @sara12345
Someone lay me out ab > m^2 -> Using the expressions we have?
I was using my letters, now they are all different.....
@shubhamsrg Use your diagram, your letters......
|dw:1350479111593:dw|
AC cosy = BC cos(2x-y) = CD cos (x-y)
This gives:|dw:1350479343315:dw|
why AC cosy = BC cos(2x-y) = CD cos (x-y) ??
Express AC, BC and CD interms of CE
you interchanged AC and BC by mistake there it seems..please confirm..
|dw:1350505825096:dw|
\((bk)^2 = b^2+CD^2 - 2.b.CD \cos C/2\) \((ak)^2 = a^2+CD^2 - 2.a.CD \cos C/2\) combining above two : \(a[b^2(k^2-1) - CD^2] = b[a^2(k^2-1)-CD^2]\) \(CD^2(b-a) = ab(k^2-1)(a-b)\) \(CD^2 = ab(1-k^2)\) \(CD < \sqrt{ab}\)
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this proof looks more elegant, it didnt even depend on angle-bisector theorem !
Is it correct?
looks flawless to me :)
it does we cancelled sinx in the first line becoz the angles are same
Oh ya..... it does.
thanks a ton again @sauravshakya and @ganeshie8 earlier i couldnt find 1 method but now i have 2 ^_^
Welcome buddy.

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