10.359. In trapezoid $ABCD$ with base lengths $AD=12 \mathrm{~cm}, BC=8 \mathrm{~cm}$, a point $M$ is taken on ray $BC$ such that $AM$ divides the trapezoid into two figures of equal area. Find $CM$.
## 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 $E N=h-y$.

Since $S_{\triangle A E D}=\frac{1}{2} S_{A B C D}$, we have $\frac{1}{2} A D \cdot E N=\frac{1}{2} \cdot \frac{A D+B C}{2} \cdot K N$, $12(h-y)=10 h, y=\frac{h}{6} \cdot \triangle A E D \sim \triangle M E C$.

![](https://cdn.mathpix.com/cropped/2024_05_21_3edb7be591bff54ce350g-0816.jpg?height=407&width=728&top_left_y=77&top_left_x=304)

Fig. 10.138

Then $\frac{M C}{A D}=\frac{K E}{E N} ; \frac{x}{12}=\frac{y}{h-y} ; x=\frac{12 y}{h-y}=\frac{12 \cdot \frac{h}{6}}{h-\frac{1}{6} h}=\frac{2 h}{\frac{5}{6} h}=2.4$.

Answer: $\quad 2.4$ cm.

![](https://cdn.mathpix.com/cropped/2024_05_21_3edb7be591bff54ce350g-0816.jpg?height=459&width=486&top_left_y=804&top_left_x=435)

Fig. 10.139