Question

In the given figure, $ DE||AC $ and $ DF||AE $ . Prove that $ \dfrac{{BF}}{{FE}} = \dfrac{{BE}}{{EC}} $

Answer 1

Hint: Line drawn parallel to one side of the triangle, intersects the other two lines and in the same ratio.
We are going to consider the given figure into two triangles which are $ \vartriangle ABC $ and $ \vartriangle AEB $ and then we are going to use the property of parallel lines inside a triangle which state “Line draw drawn parallel to one side of triangle, intersects the other two sides in distinct points, Then it divides the other 2 side in same ratio”, and then we will be able to derive two equations/ expression, which will give us what we are required to prove.

Complete step by step solution:
GIVEN:
 $ DE||AC $ and $ DF||AE $
REQUIRED TO PROOF:
 $ \dfrac{{BF}}{{FE}} = \dfrac{{BE}}{{EC}} $
PROOF:
First let us consider the $ \vartriangle ABC $
In this triangle, we consider $ DE||AC $ , by this we can apply one of the properties of parallel lines which is “Line drawn parallel to one side of a triangle, intersects the other two sides in distinct points, then it divides the other 2 side in the same ratio”. It implies to
 $ \dfrac{{BE}}{{EC}} = \dfrac{{BD}}{{DA}} $ -----------(1)
Similarly, now we are going to consider $ \vartriangle AEB $
In this triangle, we can consider $ DF||AE $ , with this we can apply the property of parallel line which says “Line drawn parallel to one side of a triangle, intersects the other two sides in distinct points, then it divides the other 2 side in the same ratio”. It implies to
 $ \dfrac{{BF}}{{FE}} = \dfrac{{BD}}{{DA}} $ ----------(2)
From the equations (1) and (2), we can conclude that
 $ \dfrac{{BF}}{{FE}} = \dfrac{{BE}}{{EC}} $
Hence proved.

Note: To prove the problem statement, we should be familiar with the properties of parallel lines; without them we cannot get the equations required to prove the required problem statement asked in the question.