To prove. If $|\overline{AC}| = |\overline{BD}|$, then $\square ABCD$ is a rectangle.
$\overline{AC}=\sqrt{(c_1-a_1)^2+(d_2-0)^2}=\sqrt{(c_1-a_1)^2+d_{2}^2}$
$\overline{BD}=\sqrt{b_{1}^2+d_{2}^2}$
If $\overline{AC}=\overline{BD}$ then $\sqrt{(c_1-a_1)^2+(d_2-0)^2}=\sqrt{b_{1}^2+d{2}^2}$ $\Rightarrow c_1-a_1=b_1$ such that $a_1<0$. If $a_1\neq 0$, then $c_1-a_1>0$ or $b_1>0$ $\Rightarrow$ $b_1>c_1>0$.
$\Rightarrow$ $\overline{AB}=(c_1-a_1)-a_1=c_1-2a_1$, such that $a_1<0$ $\Rightarrow$ $\overline{AB}>\overline{CD}$ $\Rightarrow$ $ABCD$ is not a parallelogram, which is a contradiction. Since $a_1>0$ also yields a contradiction with $a_1<0$ then it must be that $a_1=0$. Which implies that $b_1=c_1$ and $A=O$.
Therefore, $\overline{AD}^2 + \overline{AB}^2=b_{2}^2+d_{2}^2=\overline{BD}^2$. Then, by the Pythagorean Theorem, $ABCD$ is a rectangle, if $\overline{AC}=\overline{BD}$.
From the disposition of your figure you only may assume $c_1>0$, $d_2>0$, and $b_1>a_1$. From $|AC|=|BD|$ you then can infer $|c_1-a_1|=|b_1|$. How yo arrive at $a_1<0$ and later at $|AB|=c_1-2a_1$ is unclear.
Why don't you argue about angles? If the diagonals intersect at an angle $\omega$, what are the angles at the bases of the four isosceles triangles created by the diagonals?