In a parallelogram \(ABCD\), \(E\) and \(F\) are the mid-points of sides \(AB\) and \(CD\) respectively (see Fig.). Prove that the line segments \(AF\) and \(EC\) three equal parts the diagonal \(BD\).
Proof:
As, \(AB || CD\) and \(AB = CD\) () ------\((1)\)
\(AE = CF\) (Since, \(AB =CD\), given \(E\) and \(F\) are their mid-point respectively) ------\((2)\)
From (1) and (2) \(AECF\) is a
Hence, \(AF ||\)
Now taking \(∆APB\) we can see \(E\) is the mid point of side \(AB\) and
\(EQ ||\) [Given and proved]
Then by converse of Mid-point theorem, 'the line drawn through the mid-point of one side of a triangle, parallel to another side bisects the third side.'
Hence, \(Q\) is the midpoint of side \(BP\) then, \(BQ =\) -----\((3)\)
Now taking \(∆CQD\) we can see \(F\) is the mid point of side \(CD\) and \(CQ || FP\) [Given and proved]
Again by converse of Mid-point theorem, \(P\) is the mid-point of side \(DQ\).
Therefore, \(DP =\) -------\((4)\)
From (3) and (4) we conclude that, \(BQ = PQ = \)
Hence we can say that line segments \(AF\) and \(EC\) divides the diagonal \(BD\) into three equal parts.