8.4. In a right triangle $A B C$ with a right angle $B$ and angle $A$ equal to 30, the height $B D$ was drawn. Then, in triangle $B D C$, the median $D E$ was drawn, and in triangle $D E C$ - the bisector $E F$. Find the ratio of $F C$ to $A C$.
Solution: Since angle $BDC$ is a right angle, segment $DE$ is the median in the right triangle $BDC$, drawn from the vertex of the right angle. Therefore, $BE = ED = EC$. In particular, triangle $EFC$ is isosceles, so $EF$ is a median, and $2FC = DC$. Since angle $BAD$ is 30 degrees, angle $ABD$ is 60 degrees, and consequently, angle $DBC$ is 30 degrees.

Finally, in the right triangle $BDC$, angle $DBC$ is 30 degrees, so the opposite leg is half the hypotenuse, i.e., $2DC = BC$. In the right triangle $ABC$, angle $A$ is 30 degrees, so $2BC = AC$. Combining all the obtained equalities, we get $8FC = AC$, i.e., the desired ratio is $1:8$.

![](https://cdn.mathpix.com/cropped/2024_05_06_0dc9274f79a3964524a6g-1.jpg?height=751&width=494&top_left_y=1949&top_left_x=1432)

## Criteria:

Only the answer - 0 points.

Noted that $DE = EB = EC - 1$ point.

Proved that $FC = FD - 1$ point.

Applied the property of a triangle with a 30-degree angle at least once - 1 point.

All previous points are summed.