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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 $BD$?

 Aug 4, 2023
 #1
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+1

(I cheated a little on this one !!! )

 

                  C

 

          E

 

 

A     9        F   3     D          x             B

 

In triangle  ABC,   DE is paralllel to  BC

So....AE / AC = AD / AB     (1)

 

In triangle  ACD,  EF is parallel to CD

So.... AE/ AC = AF / AD      (2)

 

So  using (1) and (2)

 

AD/ AB  = AF / AD

 

So

 

AD^2 = AB * AF

 

(9+ 3)^2  = (12 + x) (9)

 

12^2  = 108 + 9x

 

144 = 108 + 9x

 

144-108 = 9x

 

36  = 9x

 

36/9  = x  = 4 =    BD

 

cool cool cool

 Aug 4, 2023

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