Problem 5. From vertex $B$ of triangle $A B C$, a line is drawn intersecting side $A C$ at point $E$. Find the height $B F$ of triangle $A B C$, given that the center of the circumscribed circle around triangle $A B C$ lies on ray $B E, A F \cdot F E=5$ and $\operatorname{ctg} \angle E B C: \operatorname{ctg} \angle B E C=3: 4$. In your answer, specify the found height, rounding it to two decimal places if necessary.
Solution. Drop a perpendicular $CD$ from point $C$ to the line $BE$. Then the ratio of the cotangents gives the proportion $BD: DE = 3: 4$. Since $DE > BD$, we have $EC > BC$. As in a right triangle the hypotenuse is greater than the leg, $BC > CF$, and therefore, point $F$ lies between points $E$ and $C$ (see Fig. 1).

By the condition, the center of the circumscribed circle lies on $BE$, hence $\angle ABE = \angle FBC$. From this, the similarity follows

$$
\begin{aligned}
& \triangle ABF \sim \triangle CBD \quad \Longrightarrow \quad \frac{BF}{AF} = \frac{BD}{CD} \\
& \triangle FBE \sim \triangle DCE \quad \Longrightarrow \quad \frac{BF}{FE} = \frac{CD}{DE}
\end{aligned}
$$

Multiplying the obtained equalities term by term, we get $\frac{BF^2}{AF \cdot FE} = \frac{BD}{DE}$. Therefore, $BF = \sqrt{AF \cdot FE \cdot \frac{BD}{DE}} = \sqrt{5 \cdot 3 / 4} = \sqrt{15} / 2$.

![](https://cdn.mathpix.com/cropped/2024_05_06_3c5c3c19db0eb78bcbc7g-18.jpg?height=551&width=1174&top_left_y=1072&top_left_x=498)

Fig. 1:

Answer: $\sqrt{15} / 2 = 1.936... \approx 1.94$.