Using Lagrange's identity, we have \begin{eqnarray} \label{1} \left(\sum_{k=1}^4 x_ky_k\right)^2 & = \left(\sum_{k=1}^4 x_k^2\right)\left(\sum_{k=1}^4 y_k^2\right) -\sum_{1 \le k < i \le 4} (x_ky_i -x_iy_k)^2\mbox{.} \end{eqnarray}We group the six squares into 3 groups of two squares and rewrite: \begin{eqnarray} \label{2} &(x_1y_2 -x_2y_1)^2 +(x_3y_4 -x_4y_3)^2 \\ \nonumber = &((x_1y_2 -x_2y_1) +(x_3y_4 -x_4y_3))^2 -2((x_1y_2 -x_2y_1)(x_3y_4 -x_4y_3)) \\ \label{3} &(x_1y_3 -x_3y_1)^2 +(x_2y_4 -x_4y_2)^2 \\ \nonumber =& ((x_1y_3 -x_3y_1) -(x_2y_4 -x_4y_2))^2 +2(x_1y_3 -x_3y_1)(x_2y_4 -x_4y_2) \\ \label{4} &(x_1y_4 -x_4y_1)^2 +(x_2y_3 -x_3y_2)^2\\ =& ((x_1y_4 -x_4y_1) +(x_2y_3 -x_3y_2))^2 -2(x_1y_4 -x_4y_1)(x_2y_3 -x_3y_2)\mbox{.} \end{eqnarray}Using \begin{eqnarray} \label{5} -2((x_1y_2 -x_2y_1)(x_3y_4 -x_4y_3)) &+2(x_1y_3 -x_3y_1)(x_2y_4 -x_4y_2) \\ \nonumber -2(x_1y_4 -x_4y_1)(x_2y_3 -x_3y_2)& =0 \end{eqnarray}we get \begin{eqnarray} \label{6} \sum_{1 \le k < i \le 4} (x_ky_i -x_iy_k)^2 & = ((x_1y_2 -x_2y_1) & +(x_3y_4 -x_4y_3))^2 \\ &+((x_1y_3 -x_3y_1) -(x_2y_4 -x_4y_2))^2 \\ \nonumber &+((x_1y_4 -x_4y_1) +(x_2y_3 -x_3y_2))^2 \end{eqnarray}by adding equations -. We put the result of equation into and get \begin{eqnarray} \left(\sum_{k=1}^4 x_ky_k\right)^2 \\ \nonumber =\left(\sum_{k=1}^4 x_k^2\right)\left(\sum_{k=1}^4 y_k^2\right) &-( (x_1y_2 -x_2y_1 +x_3y_4 -x_4y_3)^2 \\ \nonumber &-(x_1y_3 -x_3y_1 +x_4y_2 -x_2y_4 )^2 & -(x_1y_4 -x_4y_1 +x_2y_3 -x_3y_2)^2 \end{eqnarray}which is equivalent to the claimed identity.