A line drawn parallel to one side of a triangle meets the remaining two sides at distinct points. Demonstrate that the segments so formed on the two sides are in the same ratio.
Proof:
In triangles \(AED\) and \(ACB\),
\(\angle AED =\) ()
\(\angle ADE = \) ()
\(\angle EAD = \) ()
Thus, \(\Delta AED\) \(\sim \Delta ACB\) (by ).
\(\frac{AC}{AE} =\)
\(\frac{AE + EC}{AE} = \frac{AD + BD}{AD}\)
\(\frac{AE}{AE} + \frac{EC}{AE} = \frac{AD}{AD} + \frac{BD}{AD}\)
\(1 + \frac{EC}{AE} = 1 + \frac{BD}{AD}\)
\( \frac{EC}{AE} = \frac{BD}{AD}\)
Hence proved.