3. On the height $B H$ of triangle $A B C$, a point $D$ is marked. The line $A D$ intersects side $B C$ at point $E$, and the line $C D$ intersects side $A B$ at point $F$. It is known that $B H$ divides segment $F E$ in the ratio $1: 3$, counting from point $F$. Find the ratio $F H: H E$.
Answer: $1: 3$.
![](https://cdn.mathpix.com/cropped/2024_05_06_9304f7a3bc6f36cb1f9ag-14.jpg?height=482&width=1552&top_left_y=1756&top_left_x=294)

Solution 1. Let $G$ and $H$ be the projections of points $F$ and $E$ onto $AC$, respectively, $T$ be the intersection of $AE$ and $FG$, $M$ be the intersection of $CF$ and $EJ$, and $K$ be the intersection of $BH$ and $FE$ (see the left figure). Since triangles $FDT$ and $MDE$ are similar, by Thales' theorem,

$$
\frac{EM}{FT} = \frac{DE}{DT} = \frac{HJ}{GH}
$$

Triangles $AFT$ and $ABD$ are similar with a coefficient equal to the ratio of their heights. Therefore,

$$
\frac{FT}{BD} = \frac{AG}{AH} = \frac{FG}{BH}, \quad \text{that is} \quad \frac{FG}{FT} = \frac{BH}{BD}
$$

Similarly, we obtain that $\frac{EJ}{EM} = \frac{BH}{BD}$. Then

$$
\frac{FG}{FT} = \frac{EJ}{EM}, \quad \text{hence} \quad \frac{EJ}{FG} = \frac{EM}{FT} = \frac{HJ}{GH}
$$

Thus, right triangles $HEJ$ and $HFG$ are similar. Since

$$
\angle KHF = \angle HFG = \angle HEJ = \angle KHE
$$

segment $HK$ is the bisector of triangle $EHF$. Therefore,

$$
FH: HE = FK: KE = 1: 3.
$$

Solution 2. Let $K$ be the intersection of $BH$ and $EF$. Suppose the line passing through point $B$ parallel to $AC$ intersects the rays $HF$ and $HE$ at points $M$ and $N$, respectively (see the right figure). From the similarity of triangles $AFH$ and $BFM$,

$$
\frac{BM}{AH} = \frac{BF}{AF}, \quad \text{hence} \quad BM = \frac{BF \cdot AH}{AF}.
$$

The similarity of triangles $CEH$ and $BEN$ gives

$$
\frac{BN}{CH} = \frac{BE}{CE}, \quad \text{that is} \quad BN = \frac{BE \cdot CH}{CE}
$$

By Ceva's theorem,

$$
1 = \frac{BF}{AF} \cdot \frac{CE}{BE} \cdot \frac{AH}{CH} = \frac{BF \cdot AH}{AF} \cdot \frac{CE}{BE \cdot CH} = \frac{BM}{BN}
$$

Then segment $HB$ is the median and altitude of triangle $MHN$, and thus its bisector. Therefore, $HK$ is the bisector of triangle $EHF$, from which $FH: HE = FK: KE = 1: 3$.