a) Let ABCD be a quadrilateral. Consider the points and prove that EFGH is a parallelogram.

b) Let (a,A), (b,B), (c,C), (d,D) be mass-points. Consider

Under which conditions on a,b,c,d are the points E,F,G,H a parallelogram?

2. Hint: A parallelogram is a quadrilateral whose diagonals bisect each other (so the midpoint of one diagonal is equal to the midpoint of the other diagonal).

3. Originally Posted by Opalg
Hint: A parallelogram is a quadrilateral whose diagonals bisect each other (so the midpoint of one diagonal is equal to the midpoint of the other diagonal).
I really still dont know how to do this problem. I drew an arbitrary quadrilateral and used the formula A+C=B+D for a parallelogram but still cant prove it.

4. Hello, bigb!

a) Let $ABCD$ be a quadrilateral.

Consider the points: . $E \:=\:\tfrac{1}{3}(A + 2B),\;F \:=\:\tfrac{1}{3}(2B + C),\;G \:=\:\tfrac{1}{3}(C + 2D),\;H \:=\:\tfrac{1}{3}(2D+A)$

Prove that EFGH is a parallelogram.

I must assume that those equations refer to the lengths of line segments,
and that the sides are divided in the ratio 1:2 or 2:1.

The diagram looks somthing like this:
Code:
            A           E           B
o - - - - - o - - - - - o
*        *       *         *
*     *               *       *
*  *                       *     *
H o                               *   *
*    *                               * *
*         *                               o F
*              *                       *     *
*                   *               *           *
*                        *       *                 *
o - - - - - - - - - - - - - - o - - - - - - - - - - - o
D                             G                       C
Draw diagonal $BD.$

Consider $\Delta AEH\text{ and }\Delta ABD$
. . $\angle EAH \,=\,\angle BAD$
. . $AE:AB \,=\,1:3 \quad\Rightarrow\quad AE \,=\,\tfrac{1}{3}AB$
. . $AH:AD \,=\,1:3 \quad\Rightarrow\quad AH \,=\,\tfrac{1}{3}AD$
Hence: . $\Delta AEH \sim \Delta ABD$
. . $HE:DB \,=\,1:3 \quad\Rightarrow\quad HE \,=\,\tfrac{1}{3}DB$ .[1]
. . $\angle AHE\,=\,\angle ADB \quad\Rightarrow\quad HE \parallel DB$ .[2]

Consider $\Delta CFG\text{ and }\Delta CBD$
. . $\angle FCG \,=\,\angle BCD$
. . $CF:CB \,=\,1:3 \quad\Rightarrow\quad CF \,=\,\tfrac{1}{3}CB$
. . $CG:CD \,=\,1:3 \quad\Rightarrow\quad CG \,=\,\tfrac{1}{3}CD$
Hence: . $\Delta CFG \sim \Delta CBD$
. . $FG:DB \,=\,1:3\quad\Rightarrow\quad FG \,=\,\tfrac{1}{3}DB$ .[3]
. . $\angle FGC \,=\,\angle BDC \quad\Rightarrow\quad FG \parallel DB$ .[4]

From [1] and [3]: . $HE\,=\,FG$

From [2] and [4]: . $HE \parallel FG$

Theorem: If two sides of a quadrilateral are equal and parallel,
. . . . . . . the quadrilateral is a parallelogram.

Therefore, $EFGH$ is a parallelogram.