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## shubhamsrg 3 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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1. shubhamsrg

|dw:1350477585294:dw|

2. estudier

Cosine Law maybe...?

3. shubhamsrg

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..

4. estudier

|dw:1350477770138:dw|

5. estudier

bc = a^2-b^2-c^2/Cos A greater than CD^2 (from GM > CD)

6. estudier

That's on whole triangle, could maybe get more from the 2 smaller triangles..

7. shubhamsrg

am so very sorry @estudier i had mistyped the question then,,please see now..

8. estudier

OK, so ab instead of bc, same thing though...

9. shubhamsrg

okay so we have c^2 = a^2 + b^2 - 2ab cosC => ab = a^2 + b^2 - c^2 / 2cos C

10. shubhamsrg

the next step ?

11. estudier

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

12. estudier

|dw:1350478392349:dw|

13. shubhamsrg

|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)

14. shubhamsrg

we are missing the ab term..hmm..

15. estudier

Doesn't matter, the ab term is in the first equation....

16. shubhamsrg

and we had c^2 = a^2 + b^2 - 2ab cos2@ how shall we combine ?

17. estudier

Well, the other half is m^2 so may as well isolate that ...

18. estudier

Then compare the 2 sides, see how to reduce them...

19. shubhamsrg

if we subtract , we have 2cx = 2b^2 - 2 ( ab cos2@ + m(a+b)cos@ )

20. shubhamsrg

why so ?

21. sara12345

|dw:1350478727702:dw|

22. shubhamsrg

i'd second @sara12345

23. estudier

Someone lay me out ab > m^2 -> Using the expressions we have?

24. estudier

I was using my letters, now they are all different.....

25. estudier

@shubhamsrg Use your diagram, your letters......

26. sauravshakya

|dw:1350479111593:dw|

27. sauravshakya

AC cosy = BC cos(2x-y) = CD cos (x-y)

28. sauravshakya

This gives:|dw:1350479343315:dw|

29. shubhamsrg

why AC cosy = BC cos(2x-y) = CD cos (x-y) ??

30. sauravshakya

Express AC, BC and CD interms of CE

31. shubhamsrg

you interchanged AC and BC by mistake there it seems..please confirm..

32. ganeshie8

|dw:1350505825096:dw|

33. ganeshie8

$$(bk)^2 = b^2+CD^2 - 2.b.CD \cos C/2$$ $$(ak)^2 = a^2+CD^2 - 2.a.CD \cos C/2$$ combining above two : $$a[b^2(k^2-1) - CD^2] = b[a^2(k^2-1)-CD^2]$$ $$CD^2(b-a) = ab(k^2-1)(a-b)$$ $$CD^2 = ab(1-k^2)$$ $$CD < \sqrt{ab}$$

34. sauravshakya

35. ganeshie8

this proof looks more elegant, it didnt even depend on angle-bisector theorem !

36. sauravshakya

Is it correct?

37. ganeshie8

looks flawless to me :)

38. sara12345

it does we cancelled sinx in the first line becoz the angles are same

39. sauravshakya

Oh ya..... it does.

40. shubhamsrg

thanks a ton again @sauravshakya and @ganeshie8 earlier i couldnt find 1 method but now i have 2 ^_^

41. sauravshakya

Welcome buddy.

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