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
$$d^3 - (2a^2 + c^2)d - 2a^2c = 0.$$
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.