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 (see Fig. 6.14). Show that AE BF ED FC .
Step-by-Step Solution
Key Concept: When a set of parallel lines intersect two transversals, the intercepted segments on the transversals are proportional (Basic Proportionality Theorem / Thales theorem). Here AB, EF and DC are three parallel lines intersected by the transversals AD and BC.
1. Identify the parallel lines and transversals\
The three lines AB, EF and DC are parallel. The non‑parallel sides AD and BC act as transversals intersecting these parallel lines at the points:
- AD meets AB at A, EF at E and DC at D.
- BC meets AB at B, EF at F and DC at C.
2. Consider the two triangles formed by the transversals\
- Triangle \(\triangle AED\) is formed by the points where AD meets the three parallel lines.
- Triangle \(\triangle BFC\) is formed by the points where BC meets the same three parallel lines.
3. Show that the two triangles are similar\
- \(\angle AED\) and \(\angle BFC\) are both straight angles (180°) on the same line, hence they are equal.
- Since AB \(||\) DC, the angle made by AD with AB equals the angle made by AD with DC. Therefore \(\angle ADE = \angle BCF\) (alternate interior angles).
- Similarly, because AB \(||\) EF, the angle made by AD with AB equals the angle made by AD with EF, giving \(\angle AED = \angle BFC\) again.
- With two corresponding angles equal, \(\triangle AED \sim \triangle BFC\) (AA similarity).
4. Write the proportion from the similarity\
From \(\triangle AED \sim \triangle BFC\), the ratios of corresponding sides are equal:
$$\frac{AE}{ED} = \frac{BF}{FC}.$$
5. Conclusion\
Hence, the line segment \(EF\) drawn parallel to the bases of the trapezium divides the non‑parallel sides proportionally, i.e. \(AE : ED = BF : FC\).
Correct Answer: AE : ED = BF : FC