Converse of Basic Proportionality Theorem
Theorem :
If a line divides any two sides of a triangle in the same ratio, then the line is parallel to the third side.
Given :
In ∆ABC,
D and E are two points on side BC, such that \(\large \frac {AD}{DB}\) = \(\large \frac {AE}{EC}\)
To prove :
line DE || side BC
Construction:
Draw point F on side AC such that DF || side BC
Proof :
In ∆ABC
\(\large \frac {AD}{DB}\) = \(\large \frac {AE}{EC}\) … (i) [Given]
Now,
DF || side BC
∴ \(\large \frac {AD}{DB}\) = \(\large \frac {AF}{FC}\) … (ii) [By Basic Proportionality Theorem]
From equations (i) and (ii), we get,
\(\large \frac {AE}{EC}\) = \(\large \frac {AF}{FC}\)
Adding 1 on both sides, we get,
\(\large \frac {AE}{EC}\) + 1 = \(\large \frac {AF}{FC}\) + 1
∴ \(\large \frac {AE\, + \,EC}{EC}\) = \(\large \frac {AF \,+\, FC}{FC}\)
∴ \(\large \frac {AC}{EC}\) = \(\large \frac {AC}{FC}\)
∴ EC = FC
∴ Points E and F coincides
But DF || side BC …[Given]
∴ DE || side BC
Hence proved