10.358. In trapezoid $ABCD$, the bases $AD=a, BC=b$ are given. On the extension of $BC$, a point $M$ is chosen such that the line $AM$ cuts off $1/4$ of the area of the trapezoid. Find the length of the segment $CM$.
Solution.

Draw the height $K N$ of the trapezoid $A B C D$ through the point $E$ of intersection of $A M$ and $C D$ (Fig. 10.138). Let $K N=h, K E=y, C M=x$. Then $E N=h-y$. Consider the case when $S_{\triangle A E D}=\frac{1}{4} S_{A B C D}$. Then $\frac{1}{2} A D \cdot E N=\frac{1}{4} \cdot \frac{A D+B C}{2} \cdot K N . a(h-y)=\frac{1}{4}(a+b) h, 4 a h-4 a y=a h+b h$,
$4 a y=3 a h-b h, y=\frac{h(3 a-b)}{4 a}$. If $S_{A B C E}=\frac{1}{4} S_{A B C D}$, then $S_{\triangle A E D}=\frac{3}{4} S_{A B C D}$, and then $a(h-y)=3 / 4(a+b) h, 4 a h-4 a y=3 a h+3 b h, y=\frac{h(a-3 b)}{4 a}$. $\triangle A E D \sim \triangle M E C$. Then $\frac{M C}{A D}=\frac{K E}{E N} ; \frac{x}{a}=\frac{y}{h-y}$.

In the first case, when $y=\frac{h(3 a-b)}{4 a}$,

$$
x=\frac{a \cdot \frac{h(3 a-b)}{4 a}}{h-\frac{h(3 a-b)}{4 a}}=\frac{3 a-b}{4\left(1-\frac{3 a-b}{4 a}\right)}=\frac{a(3 a-b)}{a+b}
$$

In the second case, when $y=\frac{h(a-3 b)}{4 a}$,

$$
x=\frac{a \cdot \frac{h(a-3 b)}{4 a}}{h-\frac{h(a-3 b)}{4 a}}=\frac{a-3 b}{4\left(1-\frac{a-3 b}{4 a}\right)}=\frac{a(a-3 b)}{3(a+b)}
$$

Answer: $\frac{a(3 a-b)}{a+b}$ or $\frac{a(a-3 b)}{3(a+b)}$.