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In triangle ABC, points D and F are on ¯AB, and E is on ¯AC such that ¯DE¯BC and ¯EF¯CD. If AF = 7 and DF = 2, then what is BD?

 

Thank you guys!!

 Jul 26, 2024

Best Answer 

 #2
avatar+130458 
+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

 

 

cool cool cool

 Jul 26, 2024
 #1
avatar+1776 
-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:

ADAB=AEAC

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:

xx+2=77+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 EFCD implies:

DFDC=EFBC

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(with x=AD,DF=2)

We also analyze:

Using the segments, note from earlier:

From triangles ADE and proportions:

AD+DF=ABb+2+2=AC (since D is in AB)

Using our earlier derived ratios:

Since DE parallel to BC and EFCD:

Applying again:

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

The final length yields:

2 as the final answer for BD.

 Jul 26, 2024
 #2
avatar+130458 
+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

 

 

cool cool cool

CPhill Jul 26, 2024

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