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|dw:1323301015086:dw|

Given that
\[a < \frac{1}{2}(b+c)\]
Prove that
\[\angle BAC < \frac{1}{2}(\angle ABC + \angle ACB)\]

WOW

ok

i thought it's related to cosine rule?
\[a^2 = b^2 +c^2 - 2bc (\cos \angle BAC)\]

I suppose you could use that too.

\[a^2 < \frac{1}{4} (b^2 + 2bc + c^2)\]\[a < \frac{1}{2}(b+c)\]
\[a^2 < \frac{1}{4} (b+c)^2\]

hmmm - needs some thought - let me think about it for a bit...

good night

what country are you in? Im in Canada and its 8:21 pm

since each of the angles A,B,C are between 0 and 180 degrees
0

I can't think if this helps or not

you are so smart

We are ALL smart in our own ways. I strongly believe that there no such thing as a dumb person :-)

|dw:1323473308275:dw|
My teacher gave me some hints.

this proof is awesome. It does not require any trignometric identities

Let's ask myininaya when he gets on

It might be worth asking JamesJ or Zarkon as well.

i will look at the hint and see if i can understand it. back later

ok thank you for trying

so (angle B + angle C)/2 = 90 - angle A

angle A < 90 - angle A
That's what we need to prove

what you have wrote there now is so from this resulted that angle A < 45

so than angle B + angle C >= 135

and 135/2 > 45 allways

good luck bye

thanks