Geometry Help - 1 of 2

To solve the problem, we will use the properties of similar triangles, as both pairs of parallel lines indicate certain proportional relationships.

Since $DE$ is parallel to $BC$, triangles $ADE$ and $ABC$ are similar by the Basic Proportionality Theorem (also known as Thales's theorem). This implies that the ratios of corresponding segments are equal:

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

Given that:

- $AE = 7$
- $DF = 2$

Let $AD = x$. This means $AB = AD + DF = x + 2$.

Since $DE$ is parallel to $BC$, we can express $AC$ as follows:

Let $EC = y$. Then $AC = AE + EC = 7 + y$.

Now, we can set up a proportion using $AD$ and $AE$:

$$\frac{x}{x + 2} = \frac{7}{7 + y}$$

Cross-multiplying gives:

$$7(x + 2) = 7x + 7y$$

Distributing:

$$7x + 14 = 7x + 7y$$

Subtracting $7x$ from both sides:

$$14 = 7y$$

This simplifies to:

$$y = 2$$

Now that we have $EC = 2$, we can find $AC$:

$$AC = AE + EC = 7 + 2 = 9$$

Now, to find $BD$, we analyze triangle $CDF$, where $EF$ is parallel to $CD$. Similar triangles again apply, since $EF \parallel CD$ implies:

$$\frac{DF}{DC} = \frac{EF}{BC}$$

Since we need to find the length $BD$ in relationship with those defined segments, we use the original triangle's proportions.

Returning to the ratio based on the established lengths, we consider:

Let $BD = b$. Therefore:

$$AB = x + 2 \quad \text{(with } x = AD, DF = 2\text{)}$$

We also analyze:

Using the segments, note from earlier:

From triangles $ADE$ and proportions:

$$AD + DF = AB \rightarrow b + 2 + 2 = AC \text{ (since } D \text{ is in } AB \text{)}$$

Using our earlier derived ratios:

Since $DE$ parallel to $BC$ and $EF \parallel CD$:

Applying again:

We already know $DF = 2$. Thus $BD = 2$ as well by needed lengths. Thus,

The final length yields:

$$\boxed{2}$$ as the final answer for $BD$.