10.218. A line parallel to the bases of a trapezoid passes through the point of intersection of its diagonals. Find the length of the segment of this line, enclosed between the lateral sides of the trapezoid, if the bases of the trapezoid are 4 and 12 cm.
## Solution.

In trapezoid $ABCD$ (Fig. 10.28) $BC \| AD, BC=4$ cm, $AD=12$ cm, $O$ is

![](https://cdn.mathpix.com/cropped/2024_05_21_3edb7be591bff54ce350g-0728.jpg?height=404&width=723&top_left_y=109&top_left_x=109)

Fig. 10.28

![](https://cdn.mathpix.com/cropped/2024_05_21_3edb7be591bff54ce350g-0728.jpg?height=472&width=410&top_left_y=57&top_left_x=851)

Fig. 10.29

the point of intersection of the diagonals, $MN$ is the segment of the desired length. Let $h_{1}$ be the distance between the lines $BC$ and $MN$, $h_{2}$ be the distance between $MN$ and $AD$, $h$ be the distance between $BC$ and $AD$, $h=h_{1}+h_{2}$. $\triangle MBO \sim \triangle ABD$. Then $\frac{MO}{AD}=\frac{h_{1}}{h}$ (1). $\triangle OCN \sim \triangle ACD$. Then $\frac{ON}{AD}=\frac{h_{1}}{h}$. Therefore, $MO=ON=\frac{1}{2} MN$. Let $MO=x$ cm. Then from (1) we have: $\frac{x}{12}=\frac{h_{1}}{h} \cdot \triangle MAO \sim \triangle BAC$. Then $\frac{MO}{BC}=\frac{h_{2}}{h} ; \frac{x}{4}=\frac{h_{2}}{h}$. Therefore, $\frac{x}{12}+\frac{x}{4}=\frac{h_{1}}{h}+\frac{h_{2}}{h} ; \frac{x}{3}=\frac{h_{1}+h_{2}}{h} ; x=3$. Thus, $MN=6$ cm.

Answer: 6 cm.