The theorem of the two lunules is an old theorem of plane Géométrie

History

This theorem is very old: Hippocrates de Chios (- 500) studied also the duplication of the cube, i.e. $2^\; \{1/3\}$ (not to confuse with Hippocrates (- 460, - 377)). It is called also the lunules of Hippocrates .

Statement

That is to say the right-angled triangle ABC out of B and $\backslash \; mathcal\; \{C\}$ the circumscribed circle with ABC (of diameter AC).

The lunule $L\_\; \{BC\}$ is the figure formed by the half-disc of diameter BC external with the triangle ABC, to which one removes his intersection with the disc delimited by $\backslash \; mathcal\; \{C\}$.

The lunule $L\_\; \{BA\}$ is the figure formed by the half-disc of diameter BA external with the triangle ABC, to which one removes his intersection with the disc delimited by $\backslash \; mathcal\; \{C\}$.

Then the sum of the surfaces of $L\_\; \{BC\}$ and $L\_\; \{BA\}$ (in blue on the figure) is equal to the surface of the triangle ABC (in green).

Demonstration

That is to say a right-angled triangle ABC out of B

The two white small portions are the half-circle of diameter private AC of the triangle ABC. Their surface is thus $Aire\; (AC)\; -\; Surface\; (ABC)$

The two lunules are the two half-circles diameter AB and private BC of the white parts. Their surface is thus $Aire\; (AB)\; +\; Aire\; (BC)\; -\; (Surface\; (AC)\; -\; Surface\; (ABC))$. To show the theorem, it is thus enough to show that $Aire\; (AB)\; +Aire\; (BC)\; -\; Surface\; (AC)\; =0$. I.e. that the surface of the two half-discs diameter AB and BC is equal to the surface of the half-disc of diameter AC.

However the theorem of Pythagore says to us that $AC^2\; =\; AB^2+BC^2$. Thus while multiplying by $\backslash \; frac\; \{\backslash \; pi\}\; \{4\}$ one has that $\backslash \; pi\; \backslash \; left\; (\backslash \; frac\; \{AC\}\; \{2\}\; \backslash \; right)\; ^2\; =\; \backslash \; pi\; \backslash \; left\; (\backslash \; frac\; \{AB\}\; \{2\}\; \backslash \; right)\; ^2+\; \backslash \; pi\; \backslash \; left\; (\backslash \; frac\; \{BC\}\; \{2\}\; \backslash \; right)\; ^2$, which is the required equality of the surfaces.

See too

Arbelos

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