The diagram shows $28$ lattice points, each one unit from its nearest neighbors. Segment $AB$ meets segment $CD$ at $E$. Find the length of segment $AE$. $\textbf{(A)}\ \frac{4\sqrt{5}}{3} \qquad$ $\textbf{(B)}\ \frac{5\sqrt{5}}{3} \qquad$ $\textbf{(C)}\ \frac{12\sqrt{5}}{7} \qquad$ $\textbf{(D)}\ 2\sqrt{5} \qquad$ $\textbf{(E)}\ \frac{5\sqrt{65}}{9}$
Solution
Extend $DC$ to meet the horizontal line through $A$ at $F$. Since $AF\parallel DB$, $\triangle AEF\sim \triangle BED$.
Counting lattice units in the figure, $AF=5$ and $DB=4$, so $AE:BE=AF:DB=5:4$ and $AE:AB=5:9$, so $AE=\frac59 AB=\frac59 \sqrt{3^2+6^2}=\frac{5\sqrt5}3$.
