Cyclic quadrilateral in triangle In the given figure:

If $D$, $E$ and $F$ are the mid-points of sides $AB$, $AC$ and $BC$ respectively and $AG$ is perpendicular to $BC$, then prove that $DEFG$ is a cyclic quadrilateral.
 A: Assume wlog. (why?) that $F$ is between $G$ and $C$ (as is the case in your sketch).
As $D$ is the center of the circumscribed (i.e., Thales) circle for $\Delta BAG$, we have $|DG|=|DB|$. Hence $\Delta BGD$ is isosceles and so $\angle DGB=\angle CBA$ and $\angle FGD=180^\circ-\angle CBA$.
As $FE\|BA$ and $DE\|BC$, we have that $BFED$ is a parallelogram, so $\angle EDF=\angle CBA$.
It follows that $\angle FGD+\angle EDF=180^\circ$, as desired.
A: Hint:
Let M be intersection pt of DE & AG.
Then $AG = 2AM(corr. sides, \cong \triangle) $ ==> $AM = GM$
$ DE = \frac{1}{2}BC(corr. sides, \sim \triangle) $ ==> $DE = BF = CF$
(These two can easily be proved with the reference provided.)
Try to prove the following two:
$\angle ADE = \angle EFC (AA)$
$\angle ADE = \angle GDE (RHS) $
Then the conclusion will be: $\angle EFC = \angle GDE$
A: The circle with diameter $AB$ goes through $G$ (because $\angle AGB$ is a right angle). Therefore $DG=BD$.
$\triangle ABC$ and $\triangle EFC$ are similar, so $EF$ is parallel to $DB$, and hence $EF=DB$.
Therefore $DG=EF$. So $DEFG$ is a symmetric trapezium, and therefore cyclic.
Note: apparently a trapezium is called a trapezoid in North America.
