Q: 5  In a parallelogram ABCD, E and F are the mid-points of sides AB and CD respectively (see Fig. \small 8.31). Show that the line segments AF and EC trisect the diagonal BD.

            

Answers (1)

Given: In a parallelogram ABCD, E and F are the mid-points of sides AB and CD respectively

To prove: the line segments AF and EC trisect the diagonal BD.

Proof : In quadrilatweral ABCD,     

             AB=CD                             (Given)

   \frac{1}{2}AB=\frac{1}{2}CD

\Rightarrow AE=CF         (E and F are midpoints of AB and CD)

In quadrilateral AECF,     

             AE=CF                             (Given)

            AE || CF               (Opposite sides of a parallelogram)

Hence,  AECF is a parallelogram.

In \triangle DCQ,

      F is the midpoint of DC.      (given )

              FP || CQ           (AECF  is a parallelogram)

By converse of midpoint theorem,

          P is the mid point of DQ.      

         DP= PQ....................1

Similarly,

        In \triangle ABP,

      E is the midpoint of AB.      (given )

              EQ || AP          (AECF  is a parallelogram)

By converse of midpoint theorem,

          Q is the midpoint of PB.      

         OQ= QB....................2

From 1 and 2, we have 

         DP = PQ = QB.

Hence, the line segments AF and EC trisect the diagonal BD.

 

 

 

 

 

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