9.3. Perpendiculars $B E$ and $D F$, dropped from vertices $B$ and $D$ of parallelogram $A B C D$ to sides $A D$ and $B C$ respectively, divide the parallelogram into three parts of equal area. On the extension of diagonal $B D$ beyond vertex $D$, segment $D G$ is laid off, equal to segment $B D$. Line $B E$ intersects segment $A G$ at point $H$. Find the ratio $A H: H G$.
9.3. Answer: $1: 1$. Solution. By the condition $(A E \cdot B E): 2=E D \cdot B E$, hence $A E=2 E D$. Note that $A D$ is the median of triangle $A B G$. Therefore, the segment $B H$, dividing the median $A D$ in the ratio $A E: E D=2$, is also a median of triangle $A B G$.