
Re: geometry puzzle
Posted:
Oct 9, 2013 3:36 PM


> Given a triangle, with sides A, B, C, and > opposite angles a, b, c. > > Prove: if A < (B + C)/2, then a < (b + c)/2 > >  > Rich >
Hi Rich,
using the Law of Sines
A/sin(a) = B/sin(b) = C/sin(c) = 2*R,
we have
A < (B + C)/2,
or
sin(a) < (sin(b)+sin(c))/2 = sin(b+c)/2*cos(bc)/2 < sin(b+c)/2,
or
a < (b+c)/2.
Best regards, Avni

