Which area is larger?

$\angle DAB= 90$ degrees since $DC$ is tangent to the circle and $\angle DBA= 60$ degrees, we can verify that by trigonometry:

$\tan \angle DBA= \frac{opp}{adj}= \frac {\sqrt{3}r}{r} = \sqrt{3}$ , which implies that $\angle DBA= 60$ degrees.

Let $r$ be the radius of the circle and $AB = BE= r$ , then the area of triangle $ABC$ is:

$A = \frac{1}{2}bh = \frac{\sqrt(3)}{2} r^2$ .

The area of the green region is $\frac{1}{6}\pi r^2$ . Area of a sector in a circle $\frac{60}{360}\pi r^2$ .

The area of the red region is $\frac{\sqrt{3}}{2}r^2- \frac{\pi}{6}r^2$ = Area of a triangle - Area of the green sector.

Since $\frac{\pi}{6}r^2 > \frac{\sqrt{3}}{2}r^2 - \frac{\pi}{6}r^2$ . Take $\frac{\pi}{6} = 0.524 \ \text{and} \ \frac{\sqrt{3}}{2} - \frac{\pi}{6} = 0.342$ .

which means that $\frac{\pi}{6} > \frac{\sqrt{3}}{2} - \frac{\pi}{6}$ and $\frac{\pi}{6} r^2 > (\frac{\sqrt{3}}{2} - \frac{\pi}{6}) r^2$ .

This means that the green area is larger.