Answer: OD = 4.5
In \(\triangle AOB\) and \(\triangle COD\):
- \(\angle ABO \cong \angle CDO\) (alternate angles, AB ∥ DC, BD is a transversal)
- \(\angle AOB \cong \angle COD\) (vertically opposite angles)
\(\therefore \triangle AOB \sim \triangle COD\) (AA test).
\[\frac{OB}{OD} = \frac{AB}{DC}\] \[\frac{15}{OD} = \frac{20}{6}\] \[OD = \frac{15 \times 6}{20} = 4.5\]