In the diagram, triangle $A B C$ is isosceles with $A B=A C$, and $A G$ is perpendicular to $B C$. Point $D$ is the midpoint of $A B$, point $F$ is the midpoint of $A C$, and $E$ is the point of intersection of $D F$ and $A G$. What fraction of the area of $\triangle A B C$ does the shaded area represent?
(A) $\frac{1}{12}$
(B) $\frac{1}{6}$
(C) $\frac{1}{4}$
(D) $\frac{1}{10}$
(E) $\frac{1}{8}$

![](https://cdn.mathpix.com/cropped/2024_04_20_e85ef69091c57cb633f3g-130.jpg?height=363&width=374&top_left_y=472&top_left_x=1344)
Solution 1

Since $\triangle A D F$ and $\triangle A B C$ share $\angle D A F$ and

$$
\frac{A D}{A B}=\frac{A F}{A C}=\frac{1}{2}
$$

then $\triangle A D F$ is similar to $\triangle A B C$. So the ratio of their areas is the square of the ratio of their side lengths, that is

Area of $\triangle A D F=\left(\frac{1}{2}\right)^{2} \times$ Area of $\triangle A B C$.

![](https://cdn.mathpix.com/cropped/2024_04_20_688a3b42f69739f2e244g-233.jpg?height=425&width=436&top_left_y=1045&top_left_x=1430)

Also from the similar triangles, $\angle A D F=\angle A B C$, so $D F$ is parallel to $B C$. Therefore, since $A G$ is perpendicular to $B C$, then $A G$ is also perpendicular to $D E$, so $A E$ is an altitude in $\triangle A D F$. Since $\triangle A D F$ is isosceles, $D E=E F$.

Therefore,

Area of $\triangle A D F=\frac{1}{2} \times$ Area of $\triangle A D F$

$$
=\frac{1}{8} \times \text { Area of } \triangle A B C
$$

## Solution 2

Join $G$ to $D$ and $G$ to $F$. This implies that we have four identical isosceles triangles: $\triangle A D F, \triangle G D F, \triangle B D G$, and $\triangle C F G$. Each of these identical triangles has an equal area.

Thus,

$$
\text { Area of } \begin{aligned}
\triangle A E F & =\frac{1}{2}\left(\frac{1}{4} \times \text { Area of } \triangle A B C\right) \\
& =\frac{1}{8} \times \text { Area of } \triangle A B C
\end{aligned}
$$

![](https://cdn.mathpix.com/cropped/2024_04_20_688a3b42f69739f2e244g-233.jpg?height=433&width=439&top_left_y=1868&top_left_x=1426)

ANSWER: (E)