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shubhamsrg
Group Title
ABC is a triangle where CD is the angle bisector .
prove that
geometric mean of AC and BC is greater than CD
 one year ago
 one year ago
shubhamsrg Group Title
ABC is a triangle where CD is the angle bisector . prove that geometric mean of AC and BC is greater than CD
 one year ago
 one year ago

This Question is Closed

shubhamsrg Group TitleBest ResponseYou've already chosen the best response.0
dw:1350477585294:dw
 one year ago

estudier Group TitleBest ResponseYou've already chosen the best response.1
Cosine Law maybe...?
 one year ago

shubhamsrg Group TitleBest ResponseYou've already chosen the best response.0
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..
 one year ago

estudier Group TitleBest ResponseYou've already chosen the best response.1
dw:1350477770138:dw
 one year ago

estudier Group TitleBest ResponseYou've already chosen the best response.1
bc = a^2b^2c^2/Cos A greater than CD^2 (from GM > CD)
 one year ago

estudier Group TitleBest ResponseYou've already chosen the best response.1
That's on whole triangle, could maybe get more from the 2 smaller triangles..
 one year ago

shubhamsrg Group TitleBest ResponseYou've already chosen the best response.0
am so very sorry @estudier i had mistyped the question then,,please see now..
 one year ago

estudier Group TitleBest ResponseYou've already chosen the best response.1
OK, so ab instead of bc, same thing though...
 one year ago

shubhamsrg Group TitleBest ResponseYou've already chosen the best response.0
okay so we have c^2 = a^2 + b^2  2ab cosC => ab = a^2 + b^2  c^2 / 2cos C
 one year ago

shubhamsrg Group TitleBest ResponseYou've already chosen the best response.0
the next step ?
 one year ago

estudier Group TitleBest ResponseYou've already chosen the best response.1
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
 one year ago

estudier Group TitleBest ResponseYou've already chosen the best response.1
dw:1350478392349:dw
 one year ago

shubhamsrg Group TitleBest ResponseYou've already chosen the best response.0
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)
 one year ago

shubhamsrg Group TitleBest ResponseYou've already chosen the best response.0
we are missing the ab term..hmm..
 one year ago

estudier Group TitleBest ResponseYou've already chosen the best response.1
Doesn't matter, the ab term is in the first equation....
 one year ago

shubhamsrg Group TitleBest ResponseYou've already chosen the best response.0
and we had c^2 = a^2 + b^2  2ab cos2@ how shall we combine ?
 one year ago

estudier Group TitleBest ResponseYou've already chosen the best response.1
Well, the other half is m^2 so may as well isolate that ...
 one year ago

estudier Group TitleBest ResponseYou've already chosen the best response.1
Then compare the 2 sides, see how to reduce them...
 one year ago

shubhamsrg Group TitleBest ResponseYou've already chosen the best response.0
if we subtract , we have 2cx = 2b^2  2 ( ab cos2@ + m(a+b)cos@ )
 one year ago

shubhamsrg Group TitleBest ResponseYou've already chosen the best response.0
why so ?
 one year ago

sara12345 Group TitleBest ResponseYou've already chosen the best response.0
dw:1350478727702:dw
 one year ago

shubhamsrg Group TitleBest ResponseYou've already chosen the best response.0
i'd second @sara12345
 one year ago

estudier Group TitleBest ResponseYou've already chosen the best response.1
Someone lay me out ab > m^2 > Using the expressions we have?
 one year ago

estudier Group TitleBest ResponseYou've already chosen the best response.1
I was using my letters, now they are all different.....
 one year ago

estudier Group TitleBest ResponseYou've already chosen the best response.1
@shubhamsrg Use your diagram, your letters......
 one year ago

sauravshakya Group TitleBest ResponseYou've already chosen the best response.2
dw:1350479111593:dw
 one year ago

sauravshakya Group TitleBest ResponseYou've already chosen the best response.2
AC cosy = BC cos(2xy) = CD cos (xy)
 one year ago

sauravshakya Group TitleBest ResponseYou've already chosen the best response.2
This gives:dw:1350479343315:dw
 one year ago

shubhamsrg Group TitleBest ResponseYou've already chosen the best response.0
why AC cosy = BC cos(2xy) = CD cos (xy) ??
 one year ago

sauravshakya Group TitleBest ResponseYou've already chosen the best response.2
Express AC, BC and CD interms of CE
 one year ago

shubhamsrg Group TitleBest ResponseYou've already chosen the best response.0
you interchanged AC and BC by mistake there it seems..please confirm..
 one year ago

ganeshie8 Group TitleBest ResponseYou've already chosen the best response.2
dw:1350505825096:dw
 one year ago

ganeshie8 Group TitleBest ResponseYou've already chosen the best response.2
\((bk)^2 = b^2+CD^2  2.b.CD \cos C/2\) \((ak)^2 = a^2+CD^2  2.a.CD \cos C/2\) combining above two : \(a[b^2(k^21)  CD^2] = b[a^2(k^21)CD^2]\) \(CD^2(ba) = ab(k^21)(ab)\) \(CD^2 = ab(1k^2)\) \(CD < \sqrt{ab}\)
 one year ago

ganeshie8 Group TitleBest ResponseYou've already chosen the best response.2
this proof looks more elegant, it didnt even depend on anglebisector theorem !
 one year ago

sauravshakya Group TitleBest ResponseYou've already chosen the best response.2
Is it correct?
 one year ago

ganeshie8 Group TitleBest ResponseYou've already chosen the best response.2
looks flawless to me :)
 one year ago

sara12345 Group TitleBest ResponseYou've already chosen the best response.0
it does we cancelled sinx in the first line becoz the angles are same
 one year ago

sauravshakya Group TitleBest ResponseYou've already chosen the best response.2
Oh ya..... it does.
 one year ago

shubhamsrg Group TitleBest ResponseYou've already chosen the best response.0
thanks a ton again @sauravshakya and @ganeshie8 earlier i couldnt find 1 method but now i have 2 ^_^
 one year ago

sauravshakya Group TitleBest ResponseYou've already chosen the best response.2
Welcome buddy.
 one year ago
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