The diagram shows two regular pentagons ABGDE and DEFGH. [Figure] Show that the points A, E and G are collinear. Justify your answer.

1. Find the interior angle of a regular pentagon.

For a regular pentagon, each interior angle is
$\frac{(5-2)\times 180^\circ}{5}=108^\circ.$

2. Find the angle between $DE$ and $EG$ in the lower pentagon.

In regular pentagon $DEFGH$, $EF=FG$.

So triangle $EFG$ is isosceles.

Since $\angle EFG=108^\circ$,
$\angle FEG=\angle EGF=\frac{180^\circ-108^\circ}{2}=36^\circ.$

At $E$ of the lower pentagon,
$\angle DEF=108^\circ.$

Thus
$\angle DEG=\angle DEF-\angle FEG=108^\circ-36^\circ=72^\circ.$

3. Show that $\angle AEG$ is a straight angle.

In the upper regular pentagon $ABGDE$,
$\angle AED=108^\circ.$

Therefore
$\angle AEG=\angle AED+\angle DEG=108^\circ+72^\circ=180^\circ.$

4. Conclude collinearity with justification.

Since $\angle AEG=180^\circ$, $A$, $E$ and $G$ lie on the same straight line.

Hence, $A$, $E$ and $G$ are collinear.