Solution: For any triangle, the square of any side is equal to the sum of the squares of the other two sides MINUS twice the cosine of the angle between these two sides and them. If the three sides are A, B and C, and the triangle is A, B and C, then the property-A2 = B2+C2-2B C COSAB 2 = A2+C2-2A C COSB C 2 = A2+B2-2A B COSC COSC =

It is proved that the opposite sides of A, B and C in △ABC are A, B and C respectively, and a rectangular coordinate system with A as the origin and AB as the X axis is established.

Then C(bcosA, bsinA), B(c, 0),

∴a^2=|bc|^2=(bcosa-c)^2+(bsina)^2=b^2cos^2a-2bccosa+c^2+b^2sin^2a=b^2+c^2-2bccosa，

Similarly, it can be proved that B 2 = A 2+C 2-2 ACCOSB, C 2 = A 2+B 2-2 ABCOSC.