## Task 3.

Given is a triangle $A B C$ with an obtuse angle at vertex $C$. Points $E$ and $F$ are on side $\overline{A B}$ such that $|A E|=|E F|=|F B|$. On the line $B C$ there is a point $D$ such that $B C$ is perpendicular to $E D$, and $A D$ is perpendicular to $C F$.

If the angle $\varangle C F A$ is three times the angle $\varangle B D F$, determine $|D B|:|D C|$.
## First Solution.

Let $\alpha=\varangle B D F$ and $I=C F \cap A D$. Then $\varangle E F C=3 \alpha$.

Since triangle $B D E$ is a right triangle and point $F$ is the midpoint of the hypotenuse $\overline{B E}$, it follows that $\varangle F B D=\alpha$. The angle $\varangle E F D$ is an exterior angle of triangle $B D F$, so $\varangle E F D=2 \alpha$.

Now, $\varangle C F D=\varangle C F E-\varangle D F E=\alpha$. We conclude that triangle $D F C$ is isosceles.

Due to the right angles at vertices $I$ and $D$ in triangles $F I D$ and $F I A$ and in $B D E$, we conclude that $\varangle B A D=90^{\circ}-3 \alpha, \varangle F D I=\varangle B E D=90^{\circ}-\alpha$.

Now we see that triangles $A D E$ and $A F D$ and $C D F$ and $F D B$ are similar.

We calculate:

$$
\frac{|B F|}{|E D|}=\frac{|D F|}{|E D|}=\frac{|A F|}{|A D|}=\frac{|A D|}{|A E|}
$$

from which it follows that

$$
|A D|=|A E| \sqrt{2}, \quad|E D|=|A E| \frac{1}{\sqrt{2}}
$$

Applying the Pythagorean theorem to triangle $B E D$ we get

$$
|D B|=\sqrt{|B E|^{2}-|E D|^{2}}=\sqrt{\frac{7}{2}}|A E|
$$

Due to the similarity of triangles $C D F$ and $F D B$, it follows that

$$
\frac{|B D|}{|C D|}=\frac{|B D|}{|D F|} \cdot \frac{|D F|}{|C D|}=\frac{|B D|^{2}}{|D F|^{2}}=\frac{|B D|^{2}}{|B F|^{2}}=\frac{|B D|^{2}}{|A E|^{2}}=\frac{7}{2}
$$

![](https://cdn.mathpix.com/cropped/2024_05_30_81af1dcc7f43da8d3e9cg-3.jpg?height=446&width=1291&top_left_y=1870&top_left_x=385)