Finding length of line that intersects trapezoid diagonals.

Question

In trapezoid $ABCD,$ base $\overline{AB}$ has length 6, and base $\overline{CD}$ has length 18. A line passes through the intersection of the diagonals, parallel to the bases. This line intersects $\overline{AD}$ and $\overline{BC}$ at $X$ and $Y,$ respectively. Find $XY.$

I thought to try and get the value of XY I could maybe make a proportion using the median. I don't really understand how I'd go about incorporating the diagonal lengths to find the length of a line that intersects through their intersection point.

Answer 1

Let AC and BD cross at O. Then, the similar triangles lead to $\frac{XO}{AB}= \frac{XD}{DA},\> \frac{XO}{DC}= \frac{XA}{AD}$. Add up the two ratios to get

$$\frac{XO}{AB}+ \frac{XO}{DC}=1 $$ which yields $XO = \frac{AB\cdot DC}{AB+DC}=\frac92$. Likewise, $YO= \frac92$. Thus, $XY = XO +YO =9$.

Answer 2

Let the intersection of the diagonals be $O$.

Show that

1. $\frac{OA}{OC} = \frac{6}{18}$.
2. $\frac{ OY } { AB} = \frac{OC } { AC} = \frac{ 18} { 18 + 6 } $.
3. $ OY = 4.5$.
4. Similarly, $OX = 4.5$, so $XY = 9$.