Solution

Let $E$ be the intersection point of the line parallel to side $AB$, passing through point $M$, with segment $CD$. Since $ME \| AB$, it follows that $\angle MED = \angle EDA$. Moreover, since $\angle AFD = \angle EFM$ and triangle $\triangle AFD$ is isosceles, we can conclude that $\triangle EFM$ is also isosceles with $EM = FM$. Given that $M$ is the midpoint of $BC$ and $EM \| BD$, segment $EM$ is a midsegment of triangle $\triangle CDB$. Thus,

$$
\frac{BD}{FM} = \frac{BD}{EM} = 2
$$

![](https://cdn.mathpix.com/cropped/2024_05_01_4747cb56a5540358921fg-26.jpg?height=517&width=626&top_left_y=835&top_left_x=612)
Let $D$ be a point on side $A B$ of triangle $\triangle A B C$ and $F$ the intersection of $C D$ and the median $A M$. If $A F=A D$, find the ratio between $B D$ and $F M$.

![](https://cdn.mathpix.com/cropped/2024_05_01_4747cb56a5540358921fg-25.jpg?height=402&width=482&top_left_y=2152&top_left_x=884)