(!) 11. Let the bases of the trapezoid $A B C D$ be equal to $a$ and $b$ respectively, and $O$ - the centroid of the trapezoid. In what ratio does the point $O$ divide the segment connecting the midpoints of the bases of the trapezoid. (O. $117, \mathrm{~V} .125$, P. 200. )
11. Let the height of trapezoid $ABCD$ (Fig. 162) be denoted by $h$. First, we determine the distance $z$ from point $O$ to the line $AB$. Let $AB = a, CD = b$. We divide the trapezoid by the diagonal $AC$ into two triangles $ACD$ and $ACB$ with areas $S_{1}$ and $S_{2}$ and centroids $P$ and $Q$. Then $O$ is the centroid of two material points: $\left(P, S_{1}\right)$ and $\left(Q, S_{2}\right)$; the distances from points $P$ and $Q$ to the line $AB$ are $\frac{2}{3} h$ and $\frac{1}{3} h$, respectively.

![](https://cdn.mathpix.com/cropped/2024_05_21_27f86caa754616304e40g-201.jpg?height=323&width=421&top_left_y=555&top_left_x=165)

Fig. 162. Therefore,

$z=\frac{\frac{2}{3} h \cdot S_{1}+\frac{1}{3} h \cdot S_{2}}{S_{1}+S_{2}}=$ $=\frac{\frac{2}{3} h \frac{1}{2} b h+\frac{1}{3} h \cdot \frac{1}{2} a h}{\frac{a+b}{2} \cdot h}=\frac{2 b+a}{a+b} \frac{h}{3} ;$

$$
\frac{MO}{MN}=\frac{z}{h}=\frac{2 b+a}{3(a+b)}
$$

By forming the derived proportion, we get:

$$
\frac{MO}{MN-MO}=\frac{2 b+a}{3(a+b)-(2 b+a)} \text {, i.e., } \frac{MO}{ON}=\frac{2 b+a}{2 a+b} \text {. }
$$