3. On the height $B H$ of triangle $A B C$, a point $D$ is marked. Line $A D$ intersects side $B C$ at point $E$, and line $C D$ intersects side $A B$ at point $F$. Points $G$ and $J$ are the projections of points $F$ and $E$ onto side $A C$, respectively. The area of triangle $H E J$ is twice the area of triangle $H F G$. In what ratio does the height $B H$ divide the segment $F E$?
Answer: $\sqrt{2}: 1$, measured from point $E$.

Solution 1. Let $T$ be the intersection of $A E$ and $F G$, $M$ be the intersection of $C F$ and $E J$, and $K$ be the intersection of $B H$ and $F E$ (see figure). Since triangles $F D T$ and $M D E$ are similar, by Thales' theorem

$$
\frac{E M}{F T}=\frac{D E}{D T}=\frac{H J}{G H}
$$

Triangles $A F T$ and $A B D$ are similar with a coefficient equal to the ratio of their heights. Therefore,

$$
\frac{F T}{B D}=\frac{A G}{A H}=\frac{F G}{B H}, \quad \text { that is } \quad \frac{F G}{F T}=\frac{B H}{B D}
$$

Similarly, we obtain that $\frac{E J}{E M}=\frac{B H}{B D}$. Then

$$
\frac{F G}{F T}=\frac{E J}{E M}, \quad \text { hence } \frac{E J}{F G}=\frac{E M}{F T}=\frac{H J}{G H}
$$

Thus, right triangles $H E J$ and $H F G$ are similar. By Thales' theorem

$$
\frac{E K}{K F}=\frac{H J}{G H}=\sqrt{\frac{S_{H E J}}{S_{H F G}}}=\sqrt{2}
$$

![](https://cdn.mathpix.com/cropped/2024_05_06_9304f7a3bc6f36cb1f9ag-11.jpg?height=486&width=768&top_left_y=842&top_left_x=721)

Solution 2. Let $\lambda=\frac{G H}{H A}, \mu=\frac{H J}{C H}, K-$ be the intersection of $B H$ and $F E$. Note that

$$
G F=(1-\lambda) \cdot B H, \quad E J=(1-\mu) \cdot B H
$$

By Ceva's theorem

$$
1=\frac{A F}{F B} \cdot \frac{B E}{E C} \cdot \frac{C H}{H A}=\frac{1-\lambda}{\lambda} \cdot \frac{\mu}{1-\mu} \cdot \frac{C H}{H A}=\frac{1-\lambda}{1-\mu} \cdot \frac{\mu \cdot C H}{\lambda \cdot H A}=\frac{G F}{E J} \cdot \frac{H J}{G H} .
$$

Then $\frac{H J}{G H}=\frac{E J}{G F}$, which implies the similarity of right triangles $H E J$ and $H F G$. By Thales' theorem

$$
\frac{E K}{K F}=\frac{H J}{G H}=\sqrt{\frac{S_{H E J}}{S_{H F G}}}=\sqrt{2}
$$