Question

in a parallelogram ABCD, E and F are the mid-points if sides AB and CD respectively. show that the line segments AF and EC trisect the diagonal BD​

Answer 1

Step-by-step Explanation -

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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Answer 2

GIVEN:-

ABCD is a ||gm.

E and F are the mid-points of sides AB and CD respectively.

TO PROVE:-

AF and EC trisect the diagonal BD.

i.e., to prove DP=PQ=QB

PROOF:-

$\large\sf{If\:ABCD\:is\:a\:||gm}$

$\large\sf{AB=CD}$

$\large\sf{AD=BC}$

$\large\sf{AF||EC}$.........(1)

━━━━━━━━━━━━━━━

$\large\sf{AB=CD}$

$\large\sf{\frac{1}{2}AB=\frac{1}{2}CD}$

$\large\sf{AE=FC}$(E and F are mid points of AB and CD)

$\therefore$$\large\sf{AE||FC}$

.......(2)

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$\large\sf{From\:(1)\:and\:(2)}$

$\therefore$$\large\sf{AECF\:is\:a\:||gm}$

━━━━━━━━━━━━━━━

$\large\sf{In\:∆DQC,}$

F is the mid-points of side DC and PF||QC

$\therefore$P is the mid-point of DQ(Converse of mid-point theorem)

$\therefore$$\large\sf{DP=PQ}$......(3)

$\huge\sf\purple{Similarly,}$

Q is the mid-point of DP.

$\therefore$$\large\sf{PQ=QB}$.....(4)

━━━━━━━━━━━━━━━

$\large\sf{From\:(3)\:and\:(4)}$

$\large\sf{DP=PQ=QB}$

━━━━━━━━━━━━━━━

The line segment AF and AC trisect bthe diagonal BD