Oxford University MAT 2021 Solved Papers

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