#### Answer

we have a rectangle with two equal adjacent sides, that is, a square.

#### Work Step by Step

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.