10.333. Parallelogram $A B C D$, where $A B=153 \mathrm{~cm}, A D=180$ cm, $B E=135$ cm ( $B E$ - height), is divided into three equal-area figures by lines perpendicular to $A D$. At what distance from point $A$ are the points of intersection of these perpendiculars with $A D$?
## Solution.

Let $K M$ and $L H$ be the lines dividing the parallelogram $A B C D$ (Fig. 10.119) into three equal-area figures, and $D F$ be the height of the parallelogram. Then $B K=E M=H D=L F$. Let $B K=x$ cm. From $\triangle A E B\left(\angle A E B=90^{\circ}\right)$: $A E=\sqrt{A B^{2}-B E^{2}}=72$ cm. Then $A M=(x+72) \mathrm{cm}$ and $S_{A B K M}=$ $=\frac{A M+B K}{2} \cdot B E=135(x+36) \mathrm{cm}^{2} . S_{A B C D}=A D \cdot B E=180 \cdot 135 \mathrm{~cm}^{2}$. Since $S_{A B K M}=\frac{1}{3} S_{A B C D}$, then $135(x+36)=60 \cdot 135 ; x=24$. Therefore, $A M=96 \mathrm{cm}$, $A H=A D-H D=156 \mathrm{~cm}$.

Answer: 96 cm; 156 cm.