The Projection Rule

The Projection Rule. In $\triangle ABC$: (i) $a=b\cos C+c\cos B$; (ii) $b=c\cos A+a\cos C$; (iii) $c=a\cos B+b\cos A$.

Proof (of statement (i), considering all cases). Let the altitude from $A$ meet line $BC$ at $D$. $BD$ is the projection of $AB$ on $BC$, and $DC$ is the projection of $AC$ on $BC$.

Case (i): $B$ and $C$ both acute. Then $D$ lies between $B$ and $C$.

Projection of $AB$ on $BC$ = $BD=c\cos B$; projection of $AC$ on $BC$ = $DC=b\cos C$. From the figure, $a=BC=BD+DC=c\cos B+b\cos C=b\cos C+c\cos B$.

Case (ii): $B$ obtuse. Then $D$ falls outside segment $BC$, beyond $B$.

Projection of $AB$ on $BC$ = $BD=c\cos(\pi-B)=-c\cos B$; projection of $AC$ on $BC$ is still $DC=b\cos C$. From the figure, $a=BC=DC-BD=b\cos C-(-c\cos B)=b\cos C+c\cos B$.

Case (iii): $B$ right angle. Here $D$ coincides with $B$. R.H.S. $=b\cos C+c\cos B=BC+0=a=$ L.H.S.

In every case, $a=b\cos C+c\cos B$. The cases where $C$ is obtuse or a right angle are handled symmetrically. By the same argument applied to the other two vertices, $b=c\cos A+a\cos C$ and $c=a\cos B+b\cos A$. $\blacksquare$

Ex.(8) In $\triangle ABC$, prove that $(a+b)\cos C+(b+c)\cos A+(c+a)\cos B=a+b+c$. L.H.S. $=(a\cos C+b\cos C)+(b\cos A+c\cos A)+(c\cos B+a\cos B)=(a\cos C+c\cos A)+(b\cos A+a\cos B)+(c\cos B+b\cos C)$; by the Projection Rule each bracket is one full side: $=b+c+a=$ R.H.S.

Ex.(9) In $\triangle ABC$, prove that $a(\cos C-\cos B)=2(b-c)\cos^2\dfrac{A}{2}$. By the Projection Rule, $a\cos C+c\cos A=b$ and $a\cos B+b\cos A=c$, so $a\cos C=b-c\cos A$ and $a\cos B=c-b\cos A$. L.H.S. $=a\cos C-a\cos B=(b-c\cos A)-(c-b\cos A)=b-c\cos A-c+b\cos A=(b-c)+(b-c)\cos A=(b-c)(1+\cos A)=(b-c)\times2\cos^2\dfrac{A}{2}$ (using $1+\cos A=2\cos^2\frac{A}{2}$) $=2(b-c)\cos^2\dfrac{A}{2}=$ R.H.S. …