## Task B-1.7.

Quadrilateral $A B C D$ has exactly two right angles, at vertex $A$ and at vertex $C$. On the diagonal $\overline{A C}$, points $E$ and $F$ are the feet of the perpendiculars drawn from vertices $B$ and $D$ to $\overline{A C}$, respectively. If $|A E|=3,|B E|=5$ and $|C E|=7$, what is $|D F|$?
## First Solution.

![](https://cdn.mathpix.com/cropped/2024_05_30_d4daff78382705a7c34eg-07.jpg?height=594&width=668&top_left_y=1251&top_left_x=614)

Draw the diagonal $\overline{D B}$. Triangle $D A B$ is a right triangle and it holds that:

$$
\varangle B A D = \varangle B A E + \varangle E A D = 90^{\circ}
$$

Triangle $A F D$ is a right triangle and it holds that:

$$
\varangle D A F + \varangle A D F = 90^{\circ}
$$

thus, $\varangle E A B = \varangle A D F$. (The same equality can be shown using the property of angles with perpendicular legs).

Right triangles $A E B$ and $A F D$ have the same acute angle, so they are similar by the $AA$ similarity criterion. Therefore, the following ratios are equal:

$$
\frac{|A F|}{|D F|} = \frac{|B E|}{|A E|} = \frac{5}{3}, \text{ so } |A F| = \frac{5}{3}|D F| \text{. }
$$

Similarly, we can show that $\varangle C D F = \varangle B C E$.

From $\triangle C F D \sim \triangle B E C$ it follows that

$$
\frac{|C F|}{|D F|} = \frac{|B E|}{|C E|} = \frac{5}{7}, \text{ so } |C F| = \frac{5}{7}|D F| \text{. }
$$

Since $|A E| + |E C| = |A F| + |F C| = 10$, it follows that

1 point

$$
\frac{5}{3}|D F| + \frac{5}{7}|D F| = 10, \text{ or } |D F| = \frac{21}{5} = 4.2
$$