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shubhamsrg
 4 years ago
ABC is a triangle where CD is the angle bisector .
prove that
geometric mean of AC and BC is greater than CD
shubhamsrg
 4 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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shubhamsrg
 4 years ago
Best ResponseYou've already chosen the best response.0dw:1350477585294:dw

shubhamsrg
 4 years ago
Best ResponseYou've already chosen the best response.0i 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..

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0dw:1350477770138:dw

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0bc = a^2b^2c^2/Cos A greater than CD^2 (from GM > CD)

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0That's on whole triangle, could maybe get more from the 2 smaller triangles..

shubhamsrg
 4 years ago
Best ResponseYou've already chosen the best response.0am so very sorry @estudier i had mistyped the question then,,please see now..

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0OK, so ab instead of bc, same thing though...

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

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I 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

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0dw:1350478392349:dw

shubhamsrg
 4 years ago
Best ResponseYou've already chosen the best response.0dw: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)

shubhamsrg
 4 years ago
Best ResponseYou've already chosen the best response.0we are missing the ab term..hmm..

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Doesn't matter, the ab term is in the first equation....

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

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Well, the other half is m^2 so may as well isolate that ...

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Then compare the 2 sides, see how to reduce them...

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

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0dw:1350478727702:dw

shubhamsrg
 4 years ago
Best ResponseYou've already chosen the best response.0i'd second @sara12345

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Someone lay me out ab > m^2 > Using the expressions we have?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I was using my letters, now they are all different.....

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0@shubhamsrg Use your diagram, your letters......

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0dw:1350479111593:dw

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0AC cosy = BC cos(2xy) = CD cos (xy)

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0This gives:dw:1350479343315:dw

shubhamsrg
 4 years ago
Best ResponseYou've already chosen the best response.0why AC cosy = BC cos(2xy) = CD cos (xy) ??

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Express AC, BC and CD interms of CE

shubhamsrg
 4 years ago
Best ResponseYou've already chosen the best response.0you interchanged AC and BC by mistake there it seems..please confirm..

ganeshie8
 4 years ago
Best ResponseYou've already chosen the best response.2dw:1350505825096:dw

ganeshie8
 4 years ago
Best 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 : \(a[b^2(k^21)  CD^2] = b[a^2(k^21)CD^2]\) \(CD^2(ba) = ab(k^21)(ab)\) \(CD^2 = ab(1k^2)\) \(CD < \sqrt{ab}\)

ganeshie8
 4 years ago
Best ResponseYou've already chosen the best response.2this proof looks more elegant, it didnt even depend on anglebisector theorem !

ganeshie8
 4 years ago
Best ResponseYou've already chosen the best response.2looks flawless to me :)

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0it does we cancelled sinx in the first line becoz the angles are same

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