A trapezoid's bases are 2 and $3 \mathrm{~cm}$ long. The diagonals of the trapezoid intersect at a point through which a line parallel to the bases passes, intersecting the legs at points $X$ and $Y$. What is the length of $X Y$?
Let's denote the vertices of the trapezoid as $A, B, C, D$, and the intersection of the diagonals as $E$.

# 1993-04-168-1.eps 

Triangle $A B E$ is similar to triangle $C D E$ because their corresponding sides are parallel; thus,

$$
\frac{D E}{E B}=\frac{D C}{A B}=\frac{2}{3} \quad \text { or } \quad D E=\frac{2}{3} E B
$$

Since $A B\|X Y\| C D$, by the theorem of parallel intercepts,

$$
\begin{aligned}
& \frac{E Y}{D C}=\frac{E B}{D B}=\frac{E B}{\left(\frac{2}{3}+1\right) E B}=\frac{3}{5} \\
& \frac{X E}{A B}=\frac{D E}{D B}=\frac{D E}{D E+\frac{3}{2} D E}=\frac{2}{5}
\end{aligned}
$$

From this, $X E=E Y=1.2 \text{ cm}$, so $X Y=X E+E Y=2.4 \text{ cm}$.