In square $A B C D$, points $E$ and $F$ are the midpoints of sides $B C$ and $C D$ respectively. Segments $A E$ and $B F$ intersect at point $G$.

What is larger: the area of triangle $A G F$ or the area of quadrilateral $G E C F ?$
Let the area of triangle $A G F$ be denoted by $S_{1}$, and the area of quadrilateral $G E C F$ by $S_{2}$ (see the figure). If the area of the square is $S$, then

$S_{1}+S_{2}+S_{A B E}+S_{A D F}=S$. Considering that $S_{A B E}=S_{A D F}=S / 4$, we get $S_{1}+S_{2}=S / 2$. Therefore, $S_{1}-S_{2}=\left(S / 2-S_{2}\right) - S_{2}=2\left(S / 2-S_{2}\right)=2\left(S_{B C F}-S_{2}\right)>0$ (see the figure). Hence, $S_{1}>S_{2}$.

![](https://cdn.mathpix.com/cropped/2024_05_06_ccb5665c9ecd710ea159g-39.jpg?height=561&width=558&top_left_y=11&top_left_x=753)

## Answer

$S_{A F G}>S_{C E G F}$