# 3. CONDITION

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 $A H: H G$.
Solution. According to the condition $(\mathrm{AE} \cdot \mathrm{BE}): 2=\mathrm{ED} \cdot \mathrm{BE}$, hence $\mathrm{AE}=2$ ED. Note that AD is the median of triangle ABG. Therefore, the segment $\mathrm{BH}$, dividing the median $A D$ in the ratio $\mathrm{AE}: \mathrm{ED}=2$, is also a median of triangle ABG.

Answer: $1: 1$.