SAMC

Denote by $1,2,\dots ,8$ the right angled triangles as in the above figure. Applying Pythagoras theorem successively in triangles $1,2,\dots ,8$, we get: $$\begin{array}{rl}& P{A}^2+P{O}^2-O{A}^2=0\\ & P{B}^2-P{O}^2-O{B}^2=0\\ & Q{B}^2+Q{O}^2-O{B}^2=0\\ & Q{C}^2+Q{O}^2-O{C}^2=0\\ & R{C}^2+R{O}^2-O{C}^2=0\\ & R{D}^2+R{O}^2-O{D}^2=0\\ & S{D}^2+S{O}^2-O{D}^2=0\\ & S{A}^2+S{O}^2-O{A}^2=0\end{array}$$ It follows $$\begin{array}{c}P{A}^2-P{B}^2+Q{B}^2-Q{C}^2+R{C}^2-R{D}^2+S{D}^2-S{A}^2=0.\end{array}\text{\hspace{1em}(1)}$$ The last relation is equivalent to: $$\begin{array}{c}P{A}^2-(AB-PA{)}^2+Q{B}^2-(BC-QB{)}^2+R{C}^2-(CD-RC{)}^2\\ +S{D}^2-(DA-SD{)}^2=0\end{array}$$ hence $$\begin{array}{c}2(PA\cdot AB+QB\cdot BC+RC\cdot CD+SD\cdot DA)\\ =A{B}^2+B{C}^2+C{D}^2+D{A}^2\end{array}$$ and the conclusion follows.