I/3. First, let's calculate the lengths $|A D|$ and $|C D|$. Let $|A D|=v$ and $|C D|=x$. By the Pythagorean theorem, we have $v^{2}=|A C|^{2}-x^{2}=|A B|^{2}-(|B C|-x)^{2}$, from which we get $13^{2}-x^{2}=15^{2}-(14-x)^{2}$ or $13^{2}=15^{2}-14^{2}+2 \cdot 14 \cdot x$. Therefore, $x=5$ and then $v=12$.

Since $\angle B A D=\angle D E C$, triangle $E D C$ is similar to triangle $A D B$. Therefore, $\frac{|E C|}{|C D|}=\frac{|A B|}{|A D|}$ and $\angle D B A=\angle E C D$. It follows that $|E C|=\frac{15}{9} \cdot 5=\frac{25}{3}$ and $\angle F C B=\angle C B F$, so triangle $B F C$ is isosceles with vertex $B$. Therefore, $|C F|=|F B|$.

![](https://cdn.mathpix.com/cropped/2024_06_07_16ef8862fb9f11daffb9g-07.jpg?height=460&width=582&top_left_y=224&top_left_x=1274)
Let $G$ be the midpoint of $B C$. Then $F G$ is perpendicular to $B C$. Triangle $F B G$ is similar to triangle $A B D$, so $\frac{|F B|}{|B G|}=\frac{|A B|}{|B D|}$, thus $|F B|=\frac{|A B| \cdot|B G|}{|B D|}=\frac{15 \cdot 7}{9}=\frac{35}{3}$.

The length $|E F|$ is $|E F|=|C F|-|C E|=|F B|-|C E|=\frac{35}{3}-\frac{25}{3}=\frac{10}{3}$.
Triangles $A B D$ and $E C D$ are similar ..... 1 point
Triangle $A F E$ (or $B F C$) is isosceles ..... 1 point
Conclusion $F G \| A D$ ..... 1 point
Triangles $F B G$ and $A B D$ are similar ..... 1 point
Calculation of lengths $|C D|,|B D|,|G D|,|C E|,|E D|,|A D|$ ..... 2 points
Result $|E F|=\frac{10}{3}$ ..... 1 point
3. Given the lengths of the sides of triangle $ABC: |AB|=15 \text{ cm}, |BC|=14 \text{ cm}$, and $|CA|=13 \text{ cm}$. Let $D$ be the foot of the altitude from $A$ to side $BC$, and let $E$ be a point on this altitude such that $\angle BAD = \angle DEC$. The intersection of lines $AB$ and $CE$ is denoted by $F$. Calculate $|EF|$.