11.3. 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 $G$. Which is larger: the area of triangle $A G F$ or the area of quadrilateral GECF?
Answer: $S_{A F G}>S_{C E G F}$.

Solution. 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 Fig. 11.3a, b). Let the area of the square be $S$, then $S_{1}+S_{2}+S_{A B E}+S_{A D F}=S$.

Considering that $S_{A B E}=S_{A D F}=\frac{1}{4} S$, we get: $S_{1}+S_{2}=\frac{1}{2} S$. We can reason in different ways.

![](https://cdn.mathpix.com/cropped/2024_05_06_c0cc9540fd00fc9ff7cbg-1.jpg?height=386&width=380&top_left_y=2331&top_left_x=1569)

Fig. 11.3a

First method. Form and transform the difference: $S_{1}-S_{2}=\left(\frac{1}{2} S-S_{2}\right)-S_{2}=2\left(\frac{1}{4} S-S_{2}\right)=2\left(S_{B C F}-S_{2}\right)>0$ (see Fig. 11.3a). Therefore, $S_{1}>S_{2}$, that is, $S_{A F G}>S_{C E G F}$.

Second method. Let the side of the square be $2a$, that is, $S=4a^{2}$. Then $S_{1}+S_{2}=2a^{2}$. Find $S_{3}=S_{B E G}$ (see Fig. 11.3b). Note that right triangles $A B E$ and $B C F$ are equal (by two legs). Let $\angle A B E=\angle B C F=\alpha$. Then $\angle A B G=90^{\circ}-\alpha$, which means $A E \perp B F$.

From triangle $A B E: A E=a \sqrt{5}, \sin \alpha=\frac{1}{\sqrt{5}}, B G=A B \sin \alpha=\frac{2a}{\sqrt{5}}$. Then

![](https://cdn.mathpix.com/cropped/2024_05_06_c0cc9540fd00fc9ff7cbg-2.jpg?height=374&width=391&top_left_y=133&top_left_x=1552)

Fig. 11.3b $S_{3}=\frac{1}{2} B E \cdot B G \cdot \sin \alpha=\frac{1}{2} \cdot a \cdot \frac{2a}{\sqrt{5}} \cdot \frac{1}{\sqrt{5}}=\frac{a^{2}}{5}$.

Therefore, $S_{2}=S_{B C F}-S_{3}=a^{2}-\frac{a}{5}=\frac{4a^{2}}{5}$. Then $S_{1}=2a^{2}-S_{2}=\frac{6a^{2}}{5}$. Thus, $S_{1}>S_{2}$, that is, $S_{A F G}>S_{C E G F}$.

There are also reasoning methods that allow expressing $S_{1}$ and $S_{2}$ in terms of the side of the square, using similarity, Menelaus' theorem, and so on.

## Grading Criteria

+ A complete and well-justified solution is provided

$\pm$ A generally correct reasoning is provided, with minor inaccuracies or gaps

- It is used, but not proven, that $A E$ and $B F$ are perpendicular

- Only the answer is provided
- The problem is not solved or is solved incorrectly