cylinder-section when it is transferred and so arranged on the scale-beam at $\xi$ that $\xi$ is its center of gravity. And since $\chi$ is the center of gravity in the parallelogram in the half-cylinder, and $\xi$ that of the parallelogram in the cylinder-section when its position is changed, and $\xi\theta : \theta\chi$ = the parallelogram whose center of gravity is $\chi$ : the parallelogram whose center of gravity is $\xi$, then the parallelogram whose center of gravity is $\chi$ will be in equilibrium at $\theta$ with the parallelogram whose center of gravity is $\xi$. In this way it can be proved that if another straight line is drawn in the semicircle $o\pi\rho$ perpendicular to $\pi\theta$ and on this straight line a plane is constructed perpendicular to $\pi\theta$ and is produced towards both sides of the plane in which the circle $\xi o\pi\rho$ lies, then the parallelogram formed in the half-cylinder in its present position will be in equilibrium at the point $\theta$ with the parallelogram formed in the cylinder-section if this is transferred and so arranged on the scale-beam at $\xi$ that $\xi$ is its center of-gravity; therefore also all parallelograms in the half-cylinder in their present positions will be in equilibrium at the point $\theta$ with all parallelograms of the cylinder-section if they are transferred and attached to the scale-beam at the point $\xi$; consequently also the half-cylinder in its present position will be in equilibrium at the point $\theta$ with the cylinder-section if it is transferred and so arranged on the scale-beam at $\xi$ that $\xi$ is its center of gravity. \section*{Proposition XII}