Proof: By Euclid
(related to Proposition: Prop. 12.17: Construction of Polyhedron in Outer of Concentric Spheres)
 Let two spheres have been conceived about the same center, $A$.
 So, it is necessary to inscribe a polyhedral solid in the greater sphere, not touching the lesser sphere on its surface.
 Let the spheres have been cut by some plane through the center.
 So, the sections will be circles, inasmuch as a sphere is generated by the diameter remaining behind, and a semicircle being carried around [Def. 11.14] .
 And, hence, whatever position we conceive (of for) the semicircle, the plane produced through it will make a circle on the surface of the sphere.
 And (it is) clear that (it is) also a great (circle), inasmuch as the diameter of the sphere, which is also manifestly the diameter of the semicircle and the circle, is greater than all of the (other) [straight lines drawn across in the circle or the sphere [Prop. 3.15].
 Therefore, let $BCDE$ be the circle in the greater sphere, and $FGH$ the circle in the lesser sphere.
 And let two diameters of them have been drawn at right angles to one another, (namely), $BD$ and $CE$.
 And there being two circles about the same center  (namely), $BCDE$ and $FGH$  let an equilateral and evensided polygon have been inscribed in the greater circle, $BCDE$, not touching the lesser circle, $FGH$ [Prop. 12.16], of which let the sides in the quadrant $BE$ be $BK$, $KL$, $LM$, and $ME$.
 And, $KA$ being joined, let it have been drawn across to $N$.
 And let $AO$ have been set up at point $A$, at right angles to the plane of circle $BCDE$.
 And let it meet the surface of the (greater) sphere at $O$.
 And let planes have been produced through $AO$ and each of $BD$ and $KN$.
 So, according to the aforementioned (discussion), they will make great circles on the surface of the (greater) sphere.
 Let them make (great circles), of which let $BOD$ and $KON$ be semicircles on the diameters $BD$ and $KN$ (respectively).
 And since $OA$ is at right angles to the plane of circle $BCDE$, all of the planes through $OA$ are thus also at right angles to the plane of circle $BCDE$ [Prop. 11.18].
 And, hence, the semicircles $BOD$ and $KON$ are also at right angles to the plane of circle $BCDE$.
 And since semicircles $BED$, $BOD$, and $KON$ are equal  for (they are) on the equal diameters $BD$ and $KN$ [Def. 3.1]  the quadrants $BE$, $BO$, and $KO$ are also equal to one another.
 Thus, as many sides of the polygon as are in quadrant $BE$, so many are also in quadrants $BO$ and $KO$ equal to the straight lines $BK$, $KL$, $LM$, and $ME$.
 Let them have been inscribed, and let them be $BP$, $PQ$, $QR$, $RO$, $KS$, $ST$, $TU$, and $UO$.
 And let $SP$, $TQ$, and $UR$ have been joined.
 And let perpendiculars have been drawn from $P$ and $S$ to the plane of circle $BCDE$ [Prop. 11.11].
 So, they will fall on the common sections of the planes $BD$ and $KN$ (with $BCDE$), inasmuch as the planes of $BOD$ and $KON$ are also at right angles to the plane of circle $BCDE$ [Def. 11.4] .
 Let them have fallen, and let them be $PV$ and $SW$.
 And let $WV$ have been joined.
 And since $BP$ and $KS$ are equal (circumferences) having been cut off in the equal semicircles $BOD$ and $KON$ [Prop. 3.28], and $PV$ and $SW$ are perpendiculars having been drawn (from them), $PV$ is [thus] equal to $SW$, and $BV$ to $KW$ [Prop. 3.27], [Prop. 1.26].
 And the whole of $BA$ is also equal to the whole of $KA$.
 And, thus, as $BV$ is to $VA$, so $KW$ (is) to $WA$.
 $WV$ is thus parallel to $KB$ [Prop. 6.2].
 And since $PV$ and $SW$ are each at right angles to the plane of circle $BCDE$, $PV$ is thus parallel to $SW$ [Prop. 11.6].
 And it was also shown (to be) equal to it.
 And, thus, $WV$ and $SP$ are equal and parallel [Prop. 1.33].
 And since $WV$ is parallel to $SP$, but $WV$ is parallel to $KB$, $SP$ is thus also parallel to $KB$ [Prop. 11.1].
 And $BP$ and $KS$ join them.
 Thus, the quadrilateral $KBPS$ is in one plane, inasmuch as if there are two parallel straight lines, and a random point is taken on each of them, then the straight line joining the points is in the same plane as the parallel (straight lines) [Prop. 11.7].
 So, for the same (reasons), each of the quadrilaterals $SPQT$ and $TQRU$ is also in one plane.
 And triangle $URO$ is also in one plane [Prop. 11.2].
 So, if we conceive straight lines joining points $P$, $S$, $Q$, $T$, $R$, and $U$ to $A$ then some solid polyhedral figure will have been constructed between the circumferences $BO$ and $KO$, being composed of pyramids whose bases (are) the quadrilaterals $KBPS$, $SPQT$, $TQRU$, and the triangle $URO$, and apex the point $A$.
 And if we also make the same construction on each of the sides $KL$, $LM$, and $ME$, just as on $BK$, and, further, (repeat the construction) in the remaining three quadrants, then some polyhedral figure which has been inscribed in the sphere will have been constructed, being contained by pyramids whose bases (are) the aforementioned quadrilaterals, and triangle $URO$, and the (quadrilaterals and triangles) similarly arranged to them, and apex the point $A$.
 So, I say that the aforementioned polyhedron will not touch the lesser sphere on the surface on which the circle $FGH$ is (situated).
 Let the perpendicular (straight line) $AX$ have been drawn from point $A$ to the plane $KBPS$, and let it meet the plane at point $X$ [Prop. 11.11].
 And let $XB$ and $XK$ have been joined.
 And since $AX$ is at right angles to the plane of quadrilateral $KBPS$, it is thus also at right angles to all of the straight lines joined to it which are also in the plane of the quadrilateral [Def. 11.3] .
 Thus, $AX$ is at right angles to each of $BX$ and $XK$.
 And since $AB$ is equal to $AK$, the (square) on $AB$ is also equal to the (square) on $AK$.
 And the (sum of the squares) on $AX$ and $XB$ is equal to the (square) on $AB$.
 For the angle at $X$ (is) a right angle [Prop. 1.47].
 And the (sum of the squares) on $AX$ and $XK$ is equal to the (square) on $AK$ [Prop. 1.47].
 Thus, the (sum of the squares) on $AX$ and $XB$ is equal to the (sum of the squares) on $AX$ and $XK$.
 Let the (square) on $AX$ have been subtracted from both.
 Thus, the remaining (square) on $BX$ is equal to the remaining (square) on $XK$.
 Thus, $BX$ (is) equal to $XK$.
 So, similarly, we can show that the straight lines joined from $X$ to $P$ and $S$ are equal to each of $BX$ and $XK$.
 Thus, a circle drawn (in the plane of the quadrilateral) with center $X$, and radius one of $XB$ or $XK$, will also pass through $P$ and $S$, and the quadrilateral $KBPS$ will be inside the circle.
 And since $KB$ is greater than $WV$, and $WV$ (is) equal to $SP$, $KB$ (is) thus greater than $SP$.
 And $KB$ (is) equal to each of $KS$ and $BP$.
 Thus, $KS$ and $BP$ are each greater than $SP$.
 And since quadrilateral $KBPS$ is in a circle, and $KB$, $BP$, and $KS$ are equal (to one another), and $PS$ (is) less (than them), and $BX$ is the radius of the circle, the (square) on $KB$ is thus greater than double the (square) on $BX$.^{1}
 Let the perpendicular $KY$ have been drawn from $K$ to $BV$.^{2}
 And since $BD$ is less than double $DY$, and as $BD$ is to $DY$, so the (rectangle contained) by $DB$ and $BY$ (is) to the (rectangle contained) by $DY$ and $YB$  a square being described on $BY$, and a (rectangular) parallelogram (with short side equal to $BY$) completed on $YD$  the (rectangle contained) by $DB$ and $BY$ is thus also less than double the (rectangle contained) by $DY$ and $YB$.
 And, $KD$ being joined, the (rectangle contained) by $DB$ and $BY$ is equal to the (square) on $BK$, and the (rectangle contained) by $DY$ and $YB$ equal to the (square) on $KY$ [Prop. 3.31], [Prop. 6.8 corr.] 3.
 Thus, the (square) on $KB$ is less than double the (square) on $KY$.
 But, the (square) on $KB$ is greater than double the (square) on $BX$.
 Thus, the (square) on $KY$ (is) greater than the (square) on $BX$.
 And since $BA$ is equal to $KA$, the (square) on $BA$ is equal to the (square) on $AK$.
 And the (sum of the squares) on $BX$ and $XA$ is equal to the (square) on $BA$, and the (sum of the squares) on $KY$ and $YA$ (is) equal to the (square) on $KA$ [Prop. 1.47].
 Thus, the (sum of the squares) on $BX$ and $XA$ is equal to the (sum of the squares) on $KY$ and $YA$, of which the (square) on $KY$ (is) greater than the (square) on $BX$.
 Thus, the remaining (square) on $YA$ is less than the (square) on $XA$.
 Thus, $AX$ (is) greater than $AY$.
 Thus, $AX$ is much greater than $AG$.^{3}
 And $AX$ is (a perpendicular) on one of the bases of the polyhedron, and $AG$ (is a perpendicular) on the surface of the lesser sphere.
 Hence, the polyhedron will not touch the lesser sphere on its surface.
 Thus, there being two spheres about the same center, a polyhedral solid has been inscribed in the greater sphere which does not touch the lesser sphere on its surface.
 (Which is) the very thing it was required to do.
∎
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References
Adapted from (subject to copyright, with kind permission)
 Fitzpatrick, Richard: Euclid's "Elements of Geometry"
Footnotes