In trapezium ABCD, AB and DC are parallel. A straight line is drawn parallel to AB, which intersects AD and BC at points E and F respectively. Prove that: \[ AE : ED = BF : FC \]

Q.In trapezium ABCD, AB and DC are parallel. A straight line is drawn parallel to AB, which intersects AD and BC at points E and F respectively. Prove that: \[ AE : ED = BF : FC \]

Let ABCD be a trapezium where AB ∥ DC. A straight line is drawn parallel to AB, intersecting AD and BC at points E and F respectively. **To prove:** AE : ED = BF : FC **Construction:** Join points A and C, and let this line intersect EF at point G. **Proof:** In triangle ∆ADC, since DC ∥ EG, By Thales' Theorem, we get: \[ \frac{AE}{ED} = \frac{AG}{GC} \quad \text{---(i)} \] Again, in triangle ∆ACB, since AB ∥ GF, By Thales' Theorem, we get: \[ \frac{AG}{GC} = \frac{BF}{FC} \quad \text{---(ii)} \] From (i) and (ii), we conclude: \[ \frac{AE}{ED} = \frac{BF}{FC} \Rightarrow AE : ED = BF : FC \] ( proved)