## Thomas' Calculus 13th Edition

Let two side vectors be $u$ and $v$ with two diagonals $d_1=u+v$ and $d_2=u-v$ Then, $|d_1|^2=|u|^2+|v|^2+2 u \cdot v$;$|d_2|^2=|u|^2+|v|^2 -2 u \cdot v$ This implies that $|d_1|^2=|d_2|^2$ or, $2 u \cdot v-2 u \cdot v=0$ or, $|d_1|^2=|d_2|^2 \implies u \cdot v=0$ Therefore, we get $u \perp v$ Thus, the two sides are perpendicular and the parallelogram must be a rectangle.