Baltic Way shortlist

By Pythagoras, the lengths of the diagonals of quadrangle $ABCD$ are $\sqrt{d^2-a^2}$ and $\sqrt{d^2-c^2}$. Applying Ptolemaios' Theorem to the quadrilateral $ABCD$ gives $$\sqrt{d^2-a^2}\cdot \sqrt{d^2-c^2}=ab+ac,$$ which after squaring and simplifying becomes $$d^3-(2a^2+c^2)d-2a^{2}c=0.$$ Then $d=-c$ is a root of this equation, hence, $c+d$ is a positive factor of the left-hand side. Hence, the remaining factor (which is quadratic in $d$) must vanish, and we obtain $d^2=cd+2a^2$. Let $e=2d-c$. The number $c^2+8a^2=(2d-c{)}^2=e^2$ is a square, and it follows that $8a^2=e^2-c^2$. If $e$ and $c$ both were even, then by $8\mid (e^2-c^2)$ we also have $16\mid (e^2-c^2)=8a^2$ which implies $2\mid a$, a contradiction to the fact that $a$ and $c$ are relatively prime. Hence, $e$ and $c$ both must be odd. Moreover, $e$ and $c$ are obviously relatively prime. Consequently, the factors on the right-hand side of $2a^2=\frac{e-c}{2}\cdot \frac{e+c}{2}$ are relatively prime. It follows that $d=\frac{e+c}{2}$ is a perfect square or twice a perfect square.