Oxford University MAT 2021 Solved Papers

We must have $\left|\mathit{A}\mathit{B}\right|=|\mathit{B}\mathit{C}|$ so $\sqrt{{\left(\mathit{b}-\mathit{a}\right)}^2+{\left(\mathit{c}-\mathit{b}\right)}^2}=\sqrt{{\left(\mathit{c}-\mathit{b}\right)}^2+{\left(\mathit{d}-\mathit{c}\right)}^2}$.
We must also have $\left|\mathit{B}\mathit{C}\right|=|\mathit{C}\mathit{D}|$ so $\sqrt{{\left(\mathit{c}-\mathit{b}\right)}^2+{\left(\mathit{d}-\mathit{c}\right)}^2}=\sqrt{{\left(\mathit{d}-\mathit{c}\right)}^2+{\left(\mathit{a}-\mathit{d}\right)}^2}$.
These conditions are equivalent to ${\left(\mathit{b}-\mathit{a}\right)}^2={\left(\mathit{d}-\mathit{c}\right)}^2$ and ${\left(\mathit{c}-\mathit{b}\right)}^2={\left(\mathit{a}-\mathit{d}\right)}^2$ respectively.
Using the difference of two squares, the first is equivalent to $\left(\mathit{a}-\mathit{b}+\mathit{c}-\mathit{d}\right)\left(\mathit{a}-\mathit{b}-\mathit{c}+\mathit{d}\right)=0$ and the second is equivalent to $\left(\mathit{a}-\mathit{b}+\mathit{c}-\mathit{d}\right)\left(\mathit{a}+\mathit{b}-\mathit{c}-\mathit{d}\right)=0$.
In each case, we can't have both brackets equal to zero because $\mathit{c}\ne \mathit{d}$ and $\mathit{b}\ne \mathit{c}$ because the numbers are distinct. So either $\mathit{a}-\mathit{b}+\mathit{c}-\mathit{d}=0$ or both of $\mathit{a}-\mathit{b}-\mathit{c}+\mathit{d}=0$ and $\mathit{a}+\mathit{b}-\mathit{c}-\mathit{d}=0$. That second case would imply that $\mathit{a}-\mathit{c}=0$, but the numbers are distinct so that's impossible. So we're left with just the case that $\mathit{a}-\mathit{b}+\mathit{c}-\mathit{d}=0$. We can also check that $\mathit{C}\mathit{D}=\mathit{D}\mathit{A}$ in this case, because $\sqrt{{\left(\mathit{d}-\mathit{c}\right)}^2+{\left(\mathit{a}-\mathit{d}\right)}^2}=\sqrt{{\left(\mathit{a}-\mathit{d}\right)}^2+{\left(\mathit{b}-\mathit{a}\right)}^2}$ rearranges to ${\left(\mathit{d}-\mathit{c}\right)}^2={\left(\mathit{b}-\mathit{a}\right)}^2$ which is one of the equations we already had.
The answer is (d)