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|dw:1350477585294:dw|

Cosine Law maybe...?

|dw:1350477770138:dw|

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

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

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

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

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

the next step ?

|dw:1350478392349:dw|

we are missing the ab term..hmm..

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

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

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

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

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

why so ?

|dw:1350478727702:dw|

i'd second @sara12345

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

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

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

|dw:1350479111593:dw|

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

This gives:|dw:1350479343315:dw|

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

Express AC, BC and CD interms of CE

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

|dw:1350505825096:dw|

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

Is it correct?

looks flawless to me :)

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

Oh ya..... it does.

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

Welcome buddy.