23. Given a material sphere. The internal space of it is inaccessible, but constructions on its surface are possible. How, using only a compass and a ruler, to construct on paper a segment equal to the diameter of the sphere?
23. Let us draw an arbitrary circle on the surface of a sphere from any point $A$ as the center.

Take any three points $M, \Lambda^{\prime}, P$ on this circle and, measuring the distances between them with a compass, transfer these distances to a plane. We will obtain a triangle $M^{\prime} N^{\prime} P^{\prime}$ on the plane, equal to the triangle $M N P$. Describe a circle around the triangle $M^{\prime} N^{\prime} P^{\prime}$ with center $O_{1}^{\prime}$ and find the segment $M^{\prime} O_{1}^{\prime}$, equal to the perpendicular $M O_{1}$ dropped from point $M$ to the diameter $A B$ of the sphere passing through the center $O_{1}$ of the circle described from point $A$. Using the leg $M^{\prime} O_{1}^{\prime}$ and the hypotenuse $A^{\prime} M^{\prime}$, equal to the radius $A M$, construct a right triangle $M^{\prime} O_{1}^{\prime} A^{\prime}$. Draw a line through point $M^{\prime}$ perpendicular to $M^{\prime} A^{\prime}$ until it intersects the extension of the leg $A^{\prime} O_{1}^{\prime}$ at point $B^{\prime}$. The segment $A^{\prime} B^{\prime}$ will be equal to the diameter $A B$ of the sphere.