Question

In the trapezium ABCD, side AB || side DC. Diagonals AC and DB intersect each other at O. If AB = 15, DC = 10 find $\frac{OC}{OA}$.

Answer 1

$$\begin{gathered} \mathrm{In}△\mathrm{DOC}\mathrm{and}△\mathrm{BOA},\mathrm{we}\mathrm{have}, \\ \mathrm{AB}\hspace{0.17em}\parallel \mathrm{DC}\Rightarrow \angle \mathrm{BDC}=\angle \mathrm{DBA}\mathrm{and}\angle \mathrm{DCA}=\angle \mathrm{CAB}(\mathrm{Alternate}\mathrm{angles}) \\ \mathrm{Thu}s,△\mathrm{DOC}~△\mathrm{BOA}(\mathrm{AA}\mathrm{test}) \\ \therefore \frac{\mathrm{OC}}{\mathrm{OA}}=\frac{\mathrm{DC}}{\mathrm{AB}}=\frac{10}{15}=\frac{2}{3} \end{gathered}$$