geometry problem, help please!

Thank you guys !!1

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

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)

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!