## Thomas' Calculus 13th Edition

Let two side vectors be $u$ and $v$ with two diagonals $d_1=u+v$ and $d_2=u-v$ For any rectangle when $d_1 \perp d_2$, then $d_1 \cdot d_2=0$ This means that $(-v+(-u)) \cdot (-v+u)=v \cdot v-v \cdot u+u \cdot v -u \cdot u=0$ Now, $v \cdot v-v \cdot u+u \cdot v -u \cdot u=0$ or, $|v|^2-|u|^2=0$ . But $|v| + |u| \ne 0$ so, $|v|-|u|=0$ This implies that $|v|=|u|$ Hence, we have a rectangle with two equal adjacent sides, that is, a square.