Notes
two circles inscribed in triangles in a semi-circle solution

Two Circles Inscribed in Triangles in a Semi-Circle

Solution by Lengths in an Equilateral Triangle

In the above diagram, let $r$ be the radius of the semi-circle, $p$ the radius of the purple circle, and $q$ the radius of the orange circle.

The line $O D$ bisects $A C$, so angle $C \hat{O} D$ is half of angle $C \hat{O} A$. As the angles at a point on a line add up to $180^\circ$ and the interior angle of an equilateral triangle is $60^\circ$, angle $C \hat{O} A$ is $120^\circ$ so angle $C \hat{O} D$ is $60^\circ$. Then $C O E$ is half an equilateral triangle, so $O E$ is half of $O C$, which is $r$, and then the radius of the smaller circle is $q = \frac{r}{4}$.

From the relationships between the lengths in an equilateral triangle, the radius of the circle in the equilateral triangle is $\frac{1}{2\sqrt{3}}$ of the side length, which is $r$. Hence $p = \frac{1}{2\sqrt{3}} r$, and so $r = 2 \sqrt{3} p$. Then $q = \frac{2 \sqrt{3}}{4} p$ so $\pi q^2 = \frac{3}{4} \pi p^2$. Since $\pi p^2 = 8$, the area of the smaller circle is $6$.