In triangle $ABC$, points $D$ and $F$ are on $\overline{AB},$ and $E$ is on $\overline{AC}$ such that $\overline{DE}\parallel \overline{BC}$ and $\overline{EF}\parallel \overline{CD}$. If $AF = 9$ and $DF = 3$, then what is $BD$?

Why did the triangle go to therapy? Because it had so many parallel segments, it felt like it was falling apart! Alright, let's solve this geometric dilemma. Since $\overline{DE}\parallel \overline{BC}$ and $\overline{EF}\parallel \overline{CD}$, we can use similar triangles to find the value of $BD$. Notice that triangle $ADF$ is similar to triangle $CDE$. Since $AF = 9$ and $DF = 3$, we can set up the following proportion: $\frac{DF}{AF} = \frac{DE}{CD}$. Plugging in the given values, we have $\frac{3}{9} = \frac{DE}{CD}$. Simplifying, we get $\frac{1}{3} = \frac{DE}{CD}$. Since $\overline{DE}\parallel \overline{BC}$, this means that $\frac{BD}{CD} = \frac{1}{3}$ as well. Rearranging, we find that $BD = \frac{1}{3} \cdot CD$. So, to find the value of $BD$, we need to know the length of $CD$. However, that information is not given in the problem. So, unfortunately, we cannot determine the exact value of $BD$ without more information. But don't worry, triangles always have a way of working things out!

Let $BD = x$. Since $\overline{DE}\parallel \overline{BC}$, we have $\triangle ABD \sim \triangle ADE$. Therefore, we can set up the following proportion:

$$\frac{AD}{AB} = \frac{DE}{BD}$$
Since $AD = AF + FD = 9 + 3 = 12$ and $AB = AF + BD = 9 + x$, we have:
$$\frac{12}{9 + x} = \frac{DE}{x}$$
Similarly, since $\overline{EF}\parallel \overline{CD}$, we have $\triangle AEF \sim \triangle ACD$. Setting up a similar proportion, we get:
$$\frac{AE}{AC} = \frac{EF}{CD}$$
Substituting the given values $AE = AF + FE = 9 + 3 = 12$, $AC = AE + EC = 12 + x$, and $EF = DE = \frac{AC}{CD} \cdot BD$ into the proportion, we have:
$$\frac{12}{12 + x} = \frac{\frac{AC}{CD} \cdot BD}{CD} = \frac{AC}{CD}$$
Since $\frac{AC}{CD}$ appears in both proportions, we can set them equal to each other and solve for $x$:
$$\frac{12}{9 + x} = \frac{12}{12 + x}$$
Cross-multiplying, we have:
$$12(12 + x) = 12(9 + x)$$
Expanding the equation, we get:
$$144 + 12x = 108 + 12x$$
Simplifying, we have:
\begin{align*}
144 + 12x &= 108 + 12x \\
144 - 108 &= 12x - 12x \\
36 &= 0
\end{align*}
The equation $36 = 0$ is not true. Therefore, there is no solution for $x$. In other words, we cannot determine the length of $BD$ with the given information.

To find the length of $BD$, we can use similar triangles and proportionality.

Since $\overline{DE} \parallel \overline{BC}$, we have $\triangle ADE \sim \triangle ABC$ by the property of parallel lines and transversals. Thus, we have the following proportions:

$$\frac{AD}{AB} = \frac{AE}{AC} = \frac{DE}{BC}$$

Since $\overline{EF} \parallel \overline{CD}$, we can also use similarity to relate $\triangle AEF$ and $\triangle ADC$:

$$\frac{AF}{AD} = \frac{AE}{AC} = \frac{EF}{CD}$$

We are given that $AF = 9$ and $DF = 3$. Let's call $x = BD$.

Using the proportion $\frac{AF}{AD} = \frac{EF}{CD}$ and substituting the known values, we can find $CD$:

$$\frac{9}{9+3} = \frac{EF}{CD}$$
$$\frac{3}{4} = \frac{EF}{CD}$$

Since $\overline{DE} \parallel \overline{BC}$, we have $\frac{DE}{BC} = \frac{ED}{CD}$. Therefore, we can substitute $\frac{DE}{BC}$ with $\frac{ED}{CD}$ in the proportion $\frac{AD}{AB} = \frac{AE}{AC} = \frac{DE}{BC}$:

$$\frac{AD}{AB} = \frac{AE}{AC} = \frac{ED}{CD}$$

Now, we can find $AB$ and $AC$:

$$AB = AD + DB$$
$$AC = AE + EC$$

Substituting the known values $AD = 9-3 = 6$ and $AE = 9$, we get:

$$AB = 6 + x$$
$$AC = 9 + EC$$

Now, we can set up the following proportion using the similarity of $\triangle ADE$ and $\triangle ABC$:

$$\frac{6}{6 + x} = \frac{9}{9 + EC}$$

Cross-multiplying, we get:

$$6(9 + EC) = 9(6 + x)$$

Expanding the equation, we have:

$$54 + 6EC = 54 + 9x$$

Simplifying, we get:

$$6EC = 9x$$

Dividing both sides by 3, we have:

$$2EC = 3x$$

Now, we can solve for $EC$ in terms of $x$:

$$EC = \frac{3x}{2}$$

Substituting this back into the equation $AC = 9 + EC$, we have:

$$AC = 9 + \frac{3x}{2}$$

Using the proportion $\frac{9}{9+3} = \frac{EF}{CD}$ and substituting $AC = 9 + \frac{3x}{2}$, we can now solve for $CD$:

$$\frac{3}{4} = \frac{EF}{CD}$$

Cross-multiplying, we get:

$$3CD = 4EF$$

Substituting $AC = 9 + \frac{3x}{2}$ and $EC = \frac{3x}{2}$, we have:

$$CD = \frac{4EF}{3} = \frac{4 \cdot \frac{3x}{2}}{3} = \frac{2x}{2} = x$$

Therefore, $CD = x$.

Since $\overline{AB}\parallel \overline{DE}$, we have $\triangle BDF \sim \triangle CDE$ by the property of parallel lines and transversals. This gives us the following proportion:

$$\frac{BD}{DF} = \frac{CD}{DE}$$

Substituting the known values $BD = x$ and $DF = 3$ and $CD = x$, we have:

$$\frac{x}{3} = \frac{x}{x}$$

Cross-multiplying, we get:

$$x \cdot x = 3 \cdot x$$

Simplifying, we have:

$$x^2 = 3x$$

Dividing both sides by $x$, we have:

$$x = 3$$

Therefore, $BD = x = 3$.

Taking all that gibberish out of your post, leaves me with:

In triangle ABC, points D and F are on AB and E is on AC such that DE || BC and EF || CD. If AF = 9 and DF = 3, then what is BD?

I cannot reconcile your description into a viable diagram.