Question

$ABCD$ is a trapezium with $AB \parallel DC$. $E$ and $F$ are points on non-parallel sides $AD$ and $BC$ respectively such that $EF$ is parallel to $AB$. Show that $\frac{AE}{ED} = \frac{BF}{FC}$.

Accepted Answer

Join diagonal $AC$ intersecting $EF$ at point $G$. In $	riangle ADC$, $EG \parallel DC$ (since $EF \parallel AB \parallel DC$). By BPT: $$\frac{AE}{ED} = \frac{AG}{GC} \quad --- (1)$$ In $	riangle CAB$, $GF \parallel AB$. By BPT: $$\frac{AG}{GC} = \frac{BF}{FC} \quad --- (2)$$ From (1) and (2): $$\frac{AE}{ED} = \frac{BF}{FC}$$

Line parallel to bases in a trapezium divides sides proportionally

Explanation

Practise this under board exam conditions