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In the given figure:

enter image description here 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.

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    This is the nine-point circle: the midpoints and feet of the altitudes of the three sides, and the midpoints of the segments joining the orthocentre and vertices lie on a circle. – 2017-01-22

3 Answers 3

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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.

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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: $\angle EFC = \angle GDE$

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    I'm stuck at your first deduction. Why is $AG$ equal to $DG$? – 2017-01-21
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    @TonyK Sorry for the typo – 2017-01-21
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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.