`A B C D` is a parallelogram, `E\ a n d\ F` are the mid-points of `A B\ a n d\ C D` respectively. `G H` is any line intersecting `A D ,\ E F\ a n d\ B C` at `G ,\ P\ a n d\ H` respectively. Prove that `G P=P H`

Explanation

ABCD is a parallelogram, E and F are mid points of AB and DC
and a line GH intersects AD,EF and BC at G,P and H respectively.
Join GB and let it intersect EF of X.

Now since E and F are mid points of AB and DC
so`AE=EB= (1/2)AB`
and `DF=FC=(1/2)DC`

but `AB=DC`
so AEFD and BCFE are parallelogram

so `AD||EF||BC ` ....... ( 1 )

Now In `/_\ABG`

`AE=EB`and `EX||AG `
( From ( 1 ) )
so ,X is the mid point of GB (mid point theorem )

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