Got Homework?
Connect with other students for help. It's a free community.
Here's the question you clicked on:
 0 viewing
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
 2 years ago
 2 years 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
 2 years ago
 2 years ago

This Question is Closed

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

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

estudier Group TitleBest ResponseYou've already chosen the best response.1
dw:1350477770138:dw
 2 years 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)
 2 years 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..
 2 years 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..
 2 years ago

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

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

estudier Group TitleBest ResponseYou've already chosen the best response.1
dw:1350478392349:dw
 2 years 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)
 2 years ago

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

ganeshie8 Group TitleBest ResponseYou've already chosen the best response.2
dw:1350505825096:dw
 2 years 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 : \(a[b^2(k^21)  CD^2] = b[a^2(k^21)CD^2]\) \(CD^2(ba) = ab(k^21)(ab)\) \(CD^2 = ab(1k^2)\) \(CD < \sqrt{ab}\)
 2 years ago

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

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

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

sauravshakya Group TitleBest ResponseYou've already chosen the best response.2
Oh ya..... it does.
 2 years 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 ^_^
 2 years ago

sauravshakya Group TitleBest ResponseYou've already chosen the best response.2
Welcome buddy.
 2 years ago
See more questions >>>
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.