Proof - Elementar Geometry (parallelogram)

Question

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.

Answer 1

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.

Answer 2

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.

Answer 3

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):-

$HG\parallel{EF}$
$EH\parallel{FG}$
$\therefore EFGH is a parallelogram as both the pair of opposite sides are parallel.$