In Figure, `A B C D` is a parallelogram. `E\ a n d\ F` are the mid-points of the sides `A B\ a n d\ C D` respectively. Prove that the line segments `A F\ a n d\ C E` trisect (divide into three equal parts) the diagonal `B D` .

Explanation

Solution
`ABCD` is ∥gm

`AB∥CD`
`AE∥FC`
⇒`AB=CD`
`1/2 AB= 1/2 CD`

`AE=EC`
`AECF` is ∥gm

In `/_\DQC`
`F` is mid point of `DC`
`FP∥CQ`
By converse of mid point theorem `P` is mid point of `DQ`

⇒`DP=PQ` (1)
∴`AF` and `EC` bisect BD
In `/_\APB`

`E` is mid point of `AB`

`EQ∥AP`
By converse of MPT ( mid point theorem )

`Q` is mid point of `PB`

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