Given ABCD is a trapezium such that BC || AD and AB = 4 cm . If the diagonals AC and BD intersect at Ο such that AΟ/ΟC = DΟ/ΟB = 1/2

Again given the diagonals AC and BD intersect at Ο such that AΟ/ΟC = DΟ/ΟB = 1/2.

Now from the triangle ΔAOB and ΔCOD,

∠AOB = COD    (vertically opposite angle)

AO/OC = DO/OB   (Given)

from SAS congruent criterion, 

ΔAOB ≅ ΔCOD 

Now 

AO/OC = BO/OD = AB/DC   (since corresponding sides of similar traingles are preportional)

=> 1/2 = 4/DC

=> DC = 4*2

=> DC = 8 cm