10.238. The larger base of the trapezoid is twice as large as its smaller base. A line parallel to the bases is drawn through the point of intersection of the diagonals. Find the ratio of the height of each of the two resulting trapezoids to the height of the given trapezoid.
Solution.

Let in trapezoid $ABCD$ (Fig. 10.45) $AD \| BC, AD=2BC, O$ be the intersection point of $AC$ and $BD$, and $MN$ be the line mentioned in the problem. Draw through point $O$ the height $KL$ of trapezoid $ABCD$. Then $KO$ is the height of trapezoid $MBCN$ and $\triangle BOC$, and $OL$ is the height of trapezoid $AMND$ and $\triangle AOD$. $\triangle BOC \sim \triangle AOD$. Therefore, $\frac{OK}{OL}=\frac{BC}{AD}=\frac{1}{2}$. Consequently, $\frac{OK}{OL}=\frac{1}{3}, \frac{OL}{KL}=\frac{2}{3}$.

Answer: $\quad \frac{1}{3} ; \frac{2}{3}$.

![](https://cdn.mathpix.com/cropped/2024_05_21_3edb7be591bff54ce350g-0740.jpg?height=470&width=460&top_left_y=92&top_left_x=206)

Fig. 10.46

![](https://cdn.mathpix.com/cropped/2024_05_21_3edb7be591bff54ce350g-0740.jpg?height=519&width=511&top_left_y=76&top_left_x=673)

Fig. 10.47