Let $E_i$ be the foot of the altitude from $D_i$ to the side $\overline{AC}$ in the right-angled triangle $ABC$, for each $i = 0, 1, \dots, 2018$. We have $E_0 = A$, $E_{2018} = C$.

Since the lines $D_iE_i$ are parallel to $BC$, Thales' theorem asserts that
$$|E_0E_1| = |E_1E_2| = \dots = |E_{2017}E_{2018}|.$$
From here we get $|CE_i| = |AE_{2018-i}|$ for $i = 1, 2, \dots, 2017$.

The Pythagorean theorem, applied to the right-angled triangles $CD_iE_i$ and $AD_iE_i$ for $i = 1, 2, \dots, 2017$, yields
$$|CD_i|^2 = |D_iE_i|^2 + |CE_i|^2 \quad \text{and} \quad |AD_i|^2 = |D_iE_i|^2 + |AE_i|^2$$

By subtracting these two equalities, we get
$$|CD_i|^2 - |AD_i|^2 = |CE_i|^2 - |AE_i|^2, \quad i = 1, \dots, 2017,$$
and the same relation also holds for $i = 0$ and $i = 2018$.

Adding up all these relations, we get
$$\begin{aligned} & |CD_0|^2 + |CD_1|^2 + \dots + |CD_{2018}|^2 - (|AD_0|^2 - |AD_1|^2) - \dots - (|AD_{2018}|^2 \\ & = |CE_0|^2 + |CE_1|^2 + \dots + |CE_{2018}|^2 - (|AE_0|^2 - |AE_1|^2) - \dots - (|AE_{2018}|^2 \\ & = (|CE_0|^2 - |AE_{2018}|^2) + (|CE_1|^2 - |AE_{2017}|^2) + \dots + (|CE_{2018}|^2 - |AE_0|^2) = 0. \end{aligned}$$

Since $|AD_0| = 0$, this proves the claim.