## Task 3.

Let $A B C$ be a right triangle with a right angle at vertex $C$. Circle $k$ with diameter $\overline{A C}$ intersects side $\overline{A B}$ at point $D$, and the tangent to circle $k$ at point $D$ intersects side $\overline{B C}$ at point $E$. The circumcircle of triangle $C D E$ intersects side $\overline{A B}$ at points $D$ and $F$. Determine the ratio of the areas of triangles $A B C$ and $B E F$.
## Solution.

According to Thales' theorem, $\varangle CDA = 90^{\circ}$, i.e., $\overline{CD}$ is the altitude to the hypotenuse $\overline{AB}$.

Let $P$ be the midpoint of side $\overline{AC}$. Then $\overline{DP}$ is the radius of circle $k$, so it is perpendicular to the tangent $DE$. We have $\varangle PAE + \varangle PDE = 90^{\circ} + 90^{\circ} = 180^{\circ}$, which shows that quadrilateral $APDE$ is cyclic.

Thus, point $P$ lies on the circle circumscribed around triangle $CDE$, and point $F$ also lies on this circle. Therefore, we conclude that $\varangle FEC = \varangle FDC = 90^{\circ}$, and $\varangle FPC = 180^{\circ} - \varangle FEC = 90^{\circ}$. Hence, quadrilateral $PFEC$ is a rectangle.

Since $P$ is the midpoint of side $\overline{AC}$ and $PFEC$ is a rectangle, we conclude that $\overline{PF}$ and $\overline{EF}$ are midlines, i.e., $E$ and $F$ are the midpoints of segments $\overline{BC}$ and $\overline{AB}$, respectively.

Therefore, the area of triangle $BEF$ is one-fourth of the area of triangle $ABC$, i.e., the desired ratio is 4.