Let the sides of the equilateral triangle be a.
From the picture we have for the triangles Δ ACD and Δ DAB,
∠ACD =∠ DBA =60∘, AC=AB=a, AD is the common side.
So, ΔACD ≅Δ DAB. (S−A−S rule)
Then ∠CAD =∠ DAB ⇒∠FAB =∠CAE.
So, tan∠FAB =tan∠CAE
⇒√a2−d2d=√a2−(1−d)2(1−d)
or, d2{a2−(1−d)2}=(1−d)2(a2−d2)
or, a2(1−d)2−a2−d2=0
or, a2(−2d+1)=0
⇒d=12.
If d=12 then CE=BF⇒a=1 [Since in this condition BC is nothing but the shortest distance between the given parallel lines.]
For, d=12, d2−d+1=34 and 2√d2−d+13 =1.
So, option (B) is correct.