bcosC +c cosb = a

Since abc is a triangle, then we know that:

b^2 = a^2+c^2-2ac cos(b)

==> cosb=-(b^2-a^2-c^2)/2ac

and c^2 = a^2+b^2 -2ab cos(c)

==> cos (c)= -(c^2-a^2-b^2)ab

Now we need to prove that b cos(c)+ C cos(b) = a

then,

-b(c^2-a^2-b^2)/2ab -C (b^2-a^2-c^2)/2ac

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

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

= 2a^2/2a = a

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