A community for students.
Here's the question you clicked on:
 0 viewing
 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
 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

This Question is Closed

shubhamsrg
 2 years ago
Best ResponseYou've already chosen the best response.0dw:1350477585294:dw

shubhamsrg
 2 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..

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

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

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

estudier
 2 years ago
Best ResponseYou've already chosen the best response.1OK, so ab instead of bc, same thing though...

shubhamsrg
 2 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

estudier
 2 years ago
Best ResponseYou've already chosen the best response.1I 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

shubhamsrg
 2 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
 2 years ago
Best ResponseYou've already chosen the best response.0we are missing the ab term..hmm..

estudier
 2 years ago
Best ResponseYou've already chosen the best response.1Doesn't matter, the ab term is in the first equation....

shubhamsrg
 2 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 ?

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

estudier
 2 years ago
Best ResponseYou've already chosen the best response.1Then compare the 2 sides, see how to reduce them...

shubhamsrg
 2 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@ )

sara12345
 2 years ago
Best ResponseYou've already chosen the best response.0dw:1350478727702:dw

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

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

estudier
 2 years ago
Best ResponseYou've already chosen the best response.1I was using my letters, now they are all different.....

estudier
 2 years ago
Best ResponseYou've already chosen the best response.1@shubhamsrg Use your diagram, your letters......

sauravshakya
 2 years ago
Best ResponseYou've already chosen the best response.2dw:1350479111593:dw

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

sauravshakya
 2 years ago
Best ResponseYou've already chosen the best response.2This gives:dw:1350479343315:dw

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

sauravshakya
 2 years ago
Best ResponseYou've already chosen the best response.2Express AC, BC and CD interms of CE

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

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

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

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

sara12345
 2 years ago
Best ResponseYou've already chosen the best response.0it does we cancelled sinx in the first line becoz the angles are same

sauravshakya
 2 years ago
Best ResponseYou've already chosen the best response.2Oh ya..... it does.

shubhamsrg
 2 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 ^_^
Ask your own question
Sign UpFind more explanations on OpenStudy
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.