$BC = 2.1$ cm, not $BC = 2:1$. Using the formula $\frac{A_{ABC}}{A_{DEF}} = (\frac{BC}{EF})^2$, we have $$ \frac{9}{16} = \left( \frac{2.1}{EF} \right)^2 $$ Taking the square root of both sides, $$ \frac{3}{4} = \frac{2.1}{EF} $$ $$ EF = \frac{4 \times 2.1}{3} = \frac{8.4}{3} = 2.8 $$ So, $EF = 2.8$ cm.
If \( \triangle ODC \sim \triangle OBA \) and \( \angle BOC = 125^\circ \), then \( \angle DOC = ? \)