Prove that the iidentity is true in any triangle: b*cosC+c*cosB=a

We'll prove this with the help from the cosine
theorem:

b^2=c^2+a^2-2ac*cosB,  cos B =
(c^2+a^2-b^2)/2ac

c^2=a^2+b^2-2ab*cosC,  cos C =
(a^2+b^2-c^2)/2ab

Now, we'll substitute cos B and cos C, by
the found expressions, so
that:

b*cosC+c*cosB=a

b*(a^2+b^2-c^2)/2ab
+ c*(c^2+a^2-b^2)/2ac = a

After reducing the terms, we'll
obtain the irreducible quotients:

(a^2+b^2-c^2)/2a +
(c^2+a^2-b^2)/2a = a

We'll multiply the term from the right
side, by 2a:

a^2+b^2-c^2+c^2+a^2-b^2 =
2a^2

After reducing similar terms, we'll
obtain:

a^2+a^2 = 2a^2

2a^2 =
2a^2

The last result tells us that the identity,
b*cosC+c*cosB=a, is valid for any triangle.