Croatia_2018

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 $D_i{E}_i$ are parallel to $BC$, Thales' theorem asserts that $$|E_0{E}_1|=|E_1{E}_2|=\cdots =|E_{2017}{E}_{2018}|.$$ From here we get $|C{E}_i|=|A{E}_{2018-i}|$ for $i=1,2,\dots ,2017$.

The Pythagorean theorem, applied to the right-angled triangles $C{D}_i{E}_i$ and $A{D}_i{E}_i$ for $i=1,2,\dots ,2017$, yields $$|C{D}_i{|}^2=|D_i{E}_i{|}^2+|C{E}_i{|}^2\text{and}|A{D}_i{|}^2=|D_i{E}_i{|}^2+|A{E}_i{|}^2$$

By subtracting these two equalities, we get $$|C{D}_i{|}^2-|A{D}_i{|}^2=|C{E}_i{|}^2-|A{E}_i{|}^2,i=1,\dots ,2017,$$ and the same relation also holds for $i=0$ and $i=2018$.

Adding up all these relations, we get $$\begin{array}{rl}& |C{D}_0{|}^2+|C{D}_1{|}^2+\cdots +|C{D}_{2018}{|}^2-(|A{D}_0{|}^2-|A{D}_1{|}^2)-\cdots -(|A{D}_{2018}{|}^2\\ & =|C{E}_0{|}^2+|C{E}_1{|}^2+\cdots +|C{E}_{2018}{|}^2-(|A{E}_0{|}^2-|A{E}_1{|}^2)-\cdots -(|A{E}_{2018}{|}^2\\ & =(|C{E}_0{|}^2-|A{E}_{2018}{|}^2)+(|C{E}_1{|}^2-|A{E}_{2017}{|}^2)+\cdots +(|C{E}_{2018}{|}^2-|A{E}_0{|}^2)=0.\end{array}$$

Since $|A{D}_0|=0$, this proves the claim.