Questions   
Sort: 
Aug 3, 2024
 #1
avatar+936 
0

For the first problem:

 

To address the problem, we will use the properties of similar triangles since lines \( DE \) and \( EF \) are parallel to sides \( BC \) and \( CD \) of triangle \( ABC \).

Since \( DE \) is parallel to \( BC \) and \( EF \) is parallel to \( CD \), triangle \( ADF \) is similar to triangle \( ABC \).
The sides of similar triangles are proportional, so we can set up the following proportions:

\[
\frac{AD}{AB} = \frac{AF}{AC} = \frac{DF}{BC}
\]

We know that \( AF = 7 \) and \( DF = 2 \). Let \( BD = x \). The entire length \( AB \) can be expressed as:

\[
AB = AD + DF = AD + 2
\]

Let’s designate \( AD = x \). Hence, we have \( AB = x + 2 \).

Considering the structure of triangle \( ADF \) in relation to triangle \( ABC \), we also need to calculate \( AC \) in terms of \( AE \):

Since \( EF \) is parallel to \( CD \), triangles \( AEF \) and \( ACD \) are also similar, resulting in the relationship:

\[
\frac{AE}{AC} = \frac{AF}{AB} = \frac{DF}{DC}
\]

But for our immediate task of determining the length \( BD \) just using \( AF \) and \( DF \), we will not utilize this second proportion at the moment.

As both triangles \( ADF \) and \( ABC \) are similar due to the parallel lines, we can set the ratios of the line segments as follows:

\[
\frac{AD}{AB} = \frac{AF}{AC}
\]

But we know \( AB = AD + DF \), substituting this gives us:

\[
\frac{x}{x + 2} = \frac{7}{7 + 2} = \frac{7}{9}
\]

Cross-multiplying yields:

\[
9x = 7(x + 2)
\]

Expanding the right-hand side:

\[
9x = 7x + 14
\]

Subtract \( 7x \) from both sides:

\[
2x = 14
\]

Solving for \( x \):

\[
x = \frac{14}{2} = 7
\]

Thus, we find that \( AD = 7 \).

Since we established \( BD = DF = 2 \), thus:

\[
BD = 7 - 2 = 5
\]

So the length of \( BD \) is:

\[
\boxed{5}
\]

Aug 3, 2024
Aug 2, 2024

1 Online Users