Projection Rule

10.3 The Projection Rule

Theorem. In any $\triangle ABC$,

Proof. By the sine rule, $a=2R\sin A$, $b=2R\sin B$, $c=2R\sin C$. Since $A+B+C=180^\circ$, we have $A=180^\circ-(B+C)$, so $\sin A=\sin(B+C)=\sin B\cos C+\cos B\sin C$. Multiplying by $2R$,

using $2R\sin B=b$ and $2R\sin C=c$. The other two forms follow by cycling $A\to B\to C\to A$ (equivalently $a\to b\to c\to a$). $\blacksquare$

Geometric meaning. If a perpendicular is dropped from $A$ to $BC$ meeting it at $D$, then $BD=c\cos B$ and $DC=b\cos C$ (each is the projection of one side onto $BC$), and $BD+DC=a$ — hence the name. When the triangle is obtuse at $B$ or $C$, one projection becomes negative and the picture needs a signed-length convention, but the algebraic proof above via the sine rule holds unconditionally, with no case-work.

Worked example. For the $3$–$4$–$5$ right triangle with $a=5$ (opposite the right angle), $b=4$, $c=3$: here $\cos B=3/5=0.6$ and $\cos C=4/5=0.8$. Check: $b\cos C+c\cos B=4(0.8)+3(0.6)=3.2+1.8=5=a$.

The projection rule is the algebraic engine behind several standard identities, including $(b+c)\cos A+(c+a)\cos B+(a+b)\cos C=a+b+c$, obtained simply by adding all three projection equations and regrouping the cosine terms — this is Exercise 10(a), Q4.

Obtuse case in detail. If the triangle is obtuse at $C$, the foot of the perpendicular from $A$ lands outside segment $BC$, beyond $C$, and the picture $BD=BC+CD$ needs a sign flip on $CD$. The clean algebraic route via the sine rule sidesteps this entirely, which is exactly why it is preferred over the geometric picture as the proof, even though the geometric picture remains the best way to visualise the identity in the acute case. …