Therefore, a=(b+c)/(cosB+cosC)

That is acosb+acosc = b+C.

Derived from cosine theorem:

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

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

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

a^2b+bc^2-b^3+a^2c+b^2c-c^3=2b^2c+2bc^2

a^2b-bc^2-b^3+a^2c-b^2c-c^3=0

a^2(b+c)-bc(b+c)-(b+c)(b^2-bc+c^2)=0

(b+c)(a^2-bc-b^2-c^2+bc)=0

a^2-b^2-c^2=0

b^2+c^2=a^2

Therefore, triangle ABC is a right triangle.