\(ABCD\) 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 \(\frac{AE}{ED}\ =\ \frac{BF}{FC}\).
Solution:
Construction: Join \(AC\) to intersect \(EF\) at \(G\).
Given: \(AB \parallel DC\) and \(EF \parallel AB\).
As lines parallel to the same line are parallel to each other
Therefore, \(EF\parallel \) .
Consider
In which \(EG \parallel \) as \(EF \parallel \)
By, "If a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points, the other two sides are divided in the same ratio".
So, \(\frac{AE}{ED}\ = \) \( \rightarrow (1) \)
Similarly from ,
\(\frac{CG}{AG}\ = \)
Reciporcal we get
\(\frac{AG}{CG}\ = \) \( \rightarrow (2) \)
From \((1)\) and \( (2)\)
We get \(\frac{AE}{ED}\ = \)
Hence proved.