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: Identify the given information.
ABCD is a trapezium such that AB ∥ DC. Points E and F lie on sides AD and BC respectively, and the line EF ∥ AB. Since AB ∥ DC and EF ∥ AB, it follows that EF ∥ DC as well. Thus, the three lines AB, EF, and DC are all parallel to each other.

Step 2: Apply the Basic Proportionality Theorem (Intercept Theorem).
Since AB, EF, and DC are parallel lines intersecting the two transversals AD and BC, the segments formed on these transversals are proportional. According to the Basic Proportionality Theorem, if a line is drawn parallel to one side of a triangle, it divides the other two sides in the same ratio. The same idea applies here with the parallel lines cutting the sides of the trapezium.

Step 3: Write the proportional segments.
On transversal AD, the segments are AE and ED.
On transversal BC, the corresponding segments are BF and FC.

Therefore, the ratios of the corresponding segments are equal:
\[ \frac{AE}{ED} = \frac{BF}{FC} \]

Step 4: Conclusion.
Since the three lines AB, EF, and DC are parallel and they intersect the transversals AD and BC, the segments formed are proportional. Hence it is proved that
\[ \frac{AE}{ED} = \frac{BF}{FC}. \]