ABCD is a trapezium such that BC || AD and AD = 4 cm.

Given: ABCD is a trapezium in which BC||AD and AD = 4 cm

The diagonals $\mathrm{AC}$ and $\mathrm{BD}$ intersect at $\mathrm{O}$ such that $\frac{\mathrm{AO}}{\mathrm{OC}}=\frac{\mathrm{DO}}{\mathrm{OB}}=\frac{1}{2}$

To find: DC

 In ΔAOD and ΔCOB

∠OAD=∠OCB  Alternate angles∠ODA=∠OBC  Alternate angles∠AOD=∠BOC  Vertically opposite anglesSo, ∆AOD~∆COB AAA similarityNow, correponding sides of similar ∆'s are proportional.AOCO=DOBO=ADBC⇒12=ADBC⇒1

Hence the correct answer is (b)