Question

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

 

Hint:

This property is often called the Thales' Theorem for trapeziums; it essentially shows that parallel lines intercept proportional segments on transversals.

Answer 1

Step 1: Understanding the Concept:
This problem uses the Basic Proportionality Theorem (BPT). If a line is drawn parallel to one side of a triangle, it divides the other two sides in the same ratio.
Step 2: Construction:
Join $AC$ to intersect $EF$ at point $G$.
Step 3: Applying BPT in Triangles:
In $\triangle ADC$, $EG \parallel DC$ (since $EF \parallel AB$ and $AB \parallel DC$):
\[ \frac{AE}{ED} = \frac{AG}{GC} \quad \text{---(Eq. 1)} \] In $\triangle ABC$, $GF \parallel AB$ (since $EF \parallel AB$): \[ \frac{AG}{GC} = \frac{BF}{FC} \quad \text{---(Eq. 2)} \]
Step 4: Final Answer:
From (Eq. 1) and (Eq. 2), we get: \[ \frac{AE}{ED} = \frac{BF}{FC} \] Hence proved.