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

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

shubhamsrgBest 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

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

estudierBest 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

estudierBest 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

shubhamsrgBest 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

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

shubhamsrgBest 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

estudierBest 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

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

shubhamsrgBest 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

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

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

shubhamsrgBest 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

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

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

shubhamsrgBest 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

sara12345Best ResponseYou've already chosen the best response.0
dw:1350478727702:dw
 one year ago

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

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

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

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

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

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

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

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

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

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

ganeshie8Best ResponseYou've already chosen the best response.2
dw:1350505825096:dw
 one year ago

ganeshie8Best 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

ganeshie8Best ResponseYou've already chosen the best response.2
this proof looks more elegant, it didnt even depend on anglebisector theorem !
 one year ago

ganeshie8Best ResponseYou've already chosen the best response.2
looks flawless to me :)
 one year ago

sara12345Best 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

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

shubhamsrgBest 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
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.