In trapezium ABCD, AB ∥ DC, and the diagonals AC and BD meet at O…

Step 1. Similar triangles. $\angle AOB = \angle COD$ (vertically opposite), and $\angle OBA = \angle ODC$ (alternate angles, since $AB \parallel DC$). By AA, $\triangle AOB \sim \triangle COD$, with B matching D.

Step 2. The ratio. $\dfrac{OB}{OD} = \dfrac{AB}{DC} = \dfrac{10}{15} = \dfrac{2}{3}$.

Step 3. Solve. $\dfrac{6}{OD} = \dfrac{2}{3}$, so $OD = 9$ cm.

Step 4. The diagonal. $BD = OB + OD = 6 + 9 = 15$ cm.