Proof - Elementar Geometry (parallelogram) Prove that by connecting midpoints of adjacent sides of a quadrilateral we get a parallelogram.
I'm having problems with this piece of work for some time so decided to ask for help here. Though I'm not sure I translated it purely mathematically so bear that in mind.
 A: 
Let ABCD be a quadrilateral. E,F,G,H be the mid point of each sides.
Linking AC, since BE = 1/2 AB, BF = 1/2 BC, and triangle BEF and ABC shares the same angle B, by SAS for similar shapes => triangle BEF and ABC are similar. EF is parallel to AC. In a same way, HG is parallel to AC,EH is parallel to BD and GF is parallel to BD, which shows that EFGH is a parallelogram.
A: Say quadrilateral is $ABCD$ . Midpoints of $AB,BC,CD,DA$ be $P,Q,R,S$. Join $AC$. See $PQ || AC$ and also $SR||AC$ Similarly $PS||BD$ and $RQ||BD$. The reasons for these must be obvious aren't they?
So you get a parrallelogram.
A: Check this out: 
Join AC:-
In $\triangle{ABC}$, E and F are midpoints of sides AB and BC.  $\therefore EF\parallel{AC}$ (Midpoint Theorem) 
In $\triangle{ACD}$, H and H are midpoints of sides AD and DC.  $\therefore HG\parallel{AC}$ (Midpoint Theorem) 
$EF\parallel{AC}$ 
$HG\parallel{AC}$ 
$\therefore HG\parallel{EF}$ ---->(1)
Join BD:-
In $\triangle{ABD}$, E and H are midpoints of sides AB and AD  $\therefore EF\parallel{AC}$ (Midpoint Theorem) 
In $\triangle{BCD}$, F and G are midpoints of sides BC and DC.  $\therefore HG\parallel{AC}$ (Midpoint Theorem) 
$EF\parallel{AC}$ 
$HG\parallel{AC}$ 
$\therefore EH\parallel{FG}$ ---->(2)
From (1) and (2):-
$HG\parallel{EF}$ 
$EH\parallel{FG}$ 
$\therefore EFGH is a parallelogram as both the pair of opposite sides are parallel.$
