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