Q.Show that $b \cdot \cos^2 \dfrac{C}{2} + c \cdot \cos^2 \dfrac{B}{2} = S$. (In $\triangle ABC$).

Projection Rule

The Projection Rule states that each side of a triangle equals the sum of the projections of the other two sides onto it:

Geometrically, if you drop a perpendicular from $A$ onto line $BC$ landing at foot $D$, the base splits as $BD+DC=a$, and simple right-triangle trigonometry in $\triangle ABD$ and $\triangle ACD$ gives $BD=c\cos B$, $DC=b\cos C$ — hence $a=c\cos B+b\cos C$. This picture is intuitive for an acute triangle, but the same statement holds for an obtuse triangle too (one projection effectively becomes negative and 'overshoots').

The clean algebraic proof

Rather than worrying about sign cases geometrically, the tidiest proof uses the sine rule: since $A=180^\circ-(B+C)$, we have $\sin A=\sin(B+C)=\sin B\cos C+\cos B\sin C$. Multiply both sides by $2R$ and use $2R\sin A=a$, $2R\sin B=b$, $2R\sin C=c$ to get $a=b\cos C+c\cos B$ unconditionally, with no case-work on the shape of the triangle.

Why it is useful

The projection rule is less about solving a triangle (it does not, on its own, determine unknown sides/angles from given data as cleanly as the sine or cosine rule) and more about proving identities — it is the standard first move whenever a problem asks you to establish a relation purely among $a,b,c,A,B,C$ without any numerical data. A signature use: adding all three projection equations term by term and collecting the cosine coefficients gives …