In the adjoining figure, \(\frac{AD}{BD} = \frac{AE}{EC}\) and \(\angle BDE = \angle CED\), prove that \(\Delta ABC\) is an isosceles triangle.
Proof:
Given that \(\frac{AD}{BD} = \frac{AE}{EC}\)
It implies that \(DE ||BC\), by the converse of basic proportionality theorem.
\(\Rightarrow \angle ADE = \angle AED\) - - - (i)
Now, \(\angle ADE = \angle\) ( angles) - - - (ii)
\(\angle AED = \angle\) ( angles) - - - (iii)
From eqn (i), (ii) and (iii), we get
\(\angle ABC = \angle\)
Sides opposite to equal angles are equal.
\(\Rightarrow AB = \)
Therefore, \(\Delta ABC\) is an isosceles triangle.
Hence proved.