median intersection on spherical traingles

Question

I'm stuck on how to prove that the spherical lines joining the midpoint of one of the sides to the opposite vertex on a spherical triangle would all intersect at one point.

Can anyone help please?

Answer 1

Let $O$ be the center of the unit sphere $S^2$. For two arbitrary points $P$ and $Q$ on $S^2$, denote by $c(P,Q)$ the great circle on $S^2$ determined by the points $P$ and $Q$, i.e. the circle obtained by intersecting $S^2$ with the unique plane passing through the points $O, P, Q$. By $a(PQ)$ denote the arc on the great circle $c(P,Q)$ defined the points $P$ and $Q$. There are two such arcs but we choose one of them based on context.

Let $ABC$ be a spherical triangle on $S^2$, meaning that $ABC$ is the triangle obtained with vertices $A, B, C$ and great circular edges $a(AB), \, a(BC)$ and $a(CA)$. Let $L, M, N$ be the midpoints of $a(AB), \, a(BC), \, a(CA)$. Then the line $OL$ lies in the plane $OAB$ is the interior angle bisector of the angle $\angle \, AOB$. Analogously, the lines $OM$ and $ON$ lie in the planes $OBC$ and $OCA$ respectively, and are the interior angle bisector of the angles $\angle \, BOC$ and $\angle \, COA$ respectively.

Since $O$ is the center of the unit sphere $S^2$ and $A, B, C$ are points on its surface, $OA = OB = OC = 1$. Therefore, triangles $OAB, \, OBC$ and $OCA$ are isosceles and therefore the interior angle bisectors $OL, OM$ and $ON$ of these three triangles through the vertex $O$ are also orthogonal bisectors of the straight segments $AB, \, BC$ and $CA$ respectively. Let us denote the midpoints of these three segments $AB, \, BC$ and $CA$ by $L', M'$ and $N'$ respectively, noting that $L' = OL \cap AB,\, \,\, M' = OM \cap BC, \,\,\, N' = ON \cap CA.$

Look more closely at the tetrahedron $OABC$. The lines $CL', \, AM'$ and $BN'$ are the medians in the Euclidean triangle $ABC$ so they intersect at the common centroid $G' \in \text{plane}(ABC)$. This means that the line $OG'$ lies in each of the three planes $OCL' \equiv OCL , \,\, OAM' \equiv OAM$ and $OBN' \equiv OBN$, which means that these three plane intersect at the line $OG'$, i.e. $OG' = OCL\cap OAM \cap OBN$. After intersection with the unit sphere $S^2$, this latter fact yields that the three great circles $c(CL), \, c(AM)$ and $c(BN)$ intersect in the common pair of points $G$ and $G^*$ on the surfaces of $S^2$, where $\{G,G^*\} = OG'\cap S^2$. If $G$ is the point inside the spherical triangle $ABC$, then $G$ is the centroid of $ABC$ which is the common point of intersection of the median great circles $c(CL), \, c(AM)$ and $c(BN)$.