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 = 7 and DF = 2, then what is BD?

Thank you guys!!

readyformathlearning Jul 26, 2024

#2**+1 **

C

E

A F D B

7 2

Let DB = x

Since EF parallel to CD....Triangle AEF is similar to triangle ACD

AE/AC = AF/AD

AE/AC = 7/9

Since ED parallel to BC......Triangle AED is similar to triangle ACB

AE/AC = AD/AB

AE/AC = 9/ (9 + x)

So

7/9 = 9 / (9+x)

7(9+x) = 9*9

63 + 7x = 81

7x = 18

x = 18/7 = DB

CPhill Jul 26, 2024

#1**-1 **

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 \( BD \).

bader Jul 26, 2024

#2**+1 **

Best Answer

C

E

A F D B

7 2

Let DB = x

Since EF parallel to CD....Triangle AEF is similar to triangle ACD

AE/AC = AF/AD

AE/AC = 7/9

Since ED parallel to BC......Triangle AED is similar to triangle ACB

AE/AC = AD/AB

AE/AC = 9/ (9 + x)

So

7/9 = 9 / (9+x)

7(9+x) = 9*9

63 + 7x = 81

7x = 18

x = 18/7 = DB

CPhill Jul 26, 2024