The South African Mathematical Olympiad Third Round

Drop the altitude from $A$ to $BC$, and let $F$ be its foot. Furthermore, suppose that $BF=x$ (if $F$ lies on the extension of $BC$ beyond $B$, assign a negative sign to $x$) and $AF=y$. Then, by the Pythagorean theorem, $$A{B}^2=B{F}^2+A{F}^2=x^2+y^2,$$ $$A{C}^2=C{F}^2+A{F}^2=(3-x{)}^2+y^2,$$ $$A{D}^2=D{F}^2+A{F}^2=(2-x{)}^2+y^2.$$ It follows that $$\begin{array}{rl}A{B}^2+2A{C}^2-3A{D}^2& =x^2+y^2+2(3-x{)}^2+2y^2-3(2-x{)}^2-3y^2\\ & =x^2+y^2+18-12x+2x^2+2y^2-12+12x-3x^2-3y^2\\ & =6,\end{array}$$ regardless of the values of $x$ and $y$.