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+1
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avatar+4569 

What is the number of centimeters in the length of \(EF\) if \(AB\parallel CD\parallel EF\)?

 

 

 Mar 20, 2018
 #1
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0

Thank you guys !!1

 Mar 20, 2018
edited by Guest  Mar 20, 2018
 #2
avatar+8961 
+2

To avoid getting confused by the apparent right angles, I changed the picture to make BC more slanted looking.

 

 

Since alternate interior angles are congruent....

∠CAB  =  ∠ACD    and

∠ABD  =  ∠BDC

 

So by AA similarity,  △ABE  ~ △CDE

 

So if  ED = a,  then   EB = 1.5a

 

Since corresponding angles are congruent....

∠BDC  =  ∠BEF    and

∠BCD  =  ∠BFE

 

So by AA similarity,  △BCD ~ △BFE

 

FE / CD  =  BE / BD

FE / 100  =  1.5a / (1.5a + a)

FE   =   100 * 1.5a / (1.5a + a)

FE   =   60    cm

 Mar 20, 2018
 #3
avatar+111327 
+2

Note that  angle  BAE  =  angle  ECD

And  angle ABE  =  angle  EDC

So....by AA congruency, triangle ABE  is similar to triangle CDE

 

And CD : AB   =   2: 3

 

Then the height  of CDE  = CF   is 2/3 height  of  ABE  = FB

So.... CF : FB  =  2 : 3

 

And angle BFE  = angle BCD

And angle BEF   =  angle BDC

So,again, by AA congruency, triangle BDC  is similar to triangle BEF

But  there are 5 parts  of BC....and  BF  is 3/5 of these

 

So......BF  is 3/5  of  BC 

So....EF  is 3/5  of DC   =  (3/5) (100)  =    60 (cm)

 

 

cool cool cool

 Mar 20, 2018
edited by CPhill  Mar 20, 2018
edited by CPhill  Mar 20, 2018
 #4
avatar+191 
+3

This is a case of the flagpole problem which can be solved with the formula \(ab/(a + b)\)where a and b are the lengths of the sides

Therefore,  the length of \(EF\)  is \(60\)  

another way to do this w/o the formula is graphing it

 

 

Great solution, both hectictar and CPhill!

 Mar 21, 2018

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