Show that if a line is parallel to one side of a triangle, then it divides the other two sides in the same ratio.
Statement: If a line drawn parallel to one side of a triangle intersecting the other two sides in distinct points, divides the two sides in the same ratio, then it is parallel
to third side.
Proof:
Consider a triangle \(ABC\), \(D\) and \(E\) intersects \(AB\) and \(AC\) respectively.
\(DE\) is parallel to \(BC\).
Join \(BE\) and \(CD\).
Draw \(EF\) and \(DG\) perpendicular to \(AB\) and \(AC\) respectively.
Area of triangle \(= \frac{1}{2} \times bh\)
Area of \(\Delta ADE = \)
Area of \(\Delta BDE = \)
\(\frac{ar \Delta ADE}{ar \Delta BDE}\) \(=\) \(=\) \(\rightarrow (i) \)
Area of \(\Delta ADE =\)
Area of \(\Delta DEC =\)
\(\frac{ar \Delta ADE}{ar \Delta DEC}\) \(=\) \(=\) \(\rightarrow (ii) \)
From the figure, \(\Delta BDE\) and \(\Delta DEC\) lie on the same base \(DE\).
Also, \(\Delta BDE\) and \(\Delta DEC\) lie between the same parallel \(DE\) and \(BC\).
So, area of \(=\) area of
Since the area of the triangles are same, then
\(=\)
Hence proved.