Lunar Areas

Let the side lengths of the red right triangle be $a$ and $b$ as shown. Then by Pythagorean theorem , the hypotenuse of the red right triangle, which is also the diameter of the semicircle composed of the blue and red areas, is equal to $\sqrt{a^2+b^2}$ . And we have:

$\begin{aligned} {\color{#20A900}G} +\color{#3D99F6}B & = \frac 12 \times \pi \times \left( \frac a2 \right)^2 + \frac 12 \times \pi \times \left( \frac b2 \right)^2 = \boxed {\dfrac \pi 8(a^2+b^2)} & \small \color{#3D99F6} \text{where }B \text{ is the blue area.} \\ {\color{#D61F06}R} + \color{#3D99F6} B & = \frac 12 \times \pi \times \left( \frac {\sqrt{a^2+b^2}}2\right)^2 = \boxed {\dfrac \pi 8(a^2+b^2)} \\ \implies {\color{#20A900}G} +\color{#3D99F6}B & = {\color{#D61F06}R} + \color{#3D99F6} B \end{aligned}$

Therefore $\boxed{{\color{#20A900}G} = {\color{#D61F06}R}}$ .

Let the blue area be $B$

By Pythagorean theorem, the area of the two small semicircles equal to the area of the big semicircle.

$G+B=R+B,G=R$

the above is hippocrates theorem

the below is proof