A nice analogy in Spherical Geometry with a nicer application to Ceva

It is well known that the Law of sines has a spherical counterpart. But from the Wiki article, we can infer something even better. That is,

$\displaystyle V=\frac{1}{6}\sin{a}\sin{b}\sin{C}=\frac{1}{6}\sin{b}\sin{c}\sin{A}=\frac{1}{6}\sin{c}\sin{a}\sin{B}$

where $V$ is the volume of tetrahedron formed by the vertices of the spherical triangle along with the origin. The above can be re-written as

$\displaystyle V=\frac{1}{6}\sin{a}\sin{h_a}=\frac{1}{6}\sin{b}\sin{h_b}=\frac{1}{6}\sin{c}\sin{h_c}$

where $h_a$ is the altitude passing through $A$ and so on. This way, we can see the volume described above plays a role very similar to that of area in plane geometry. From this, we note that for two spherical triangles with the same height,

$V_1/V_2=\sin{b_1}/\sin{b_2}$

This fact can be used to prove Ceva's theorem, a powerful theorem in plane geometry. Though it is well known that Ceva's theorem also works for spherical geometry, I was not able to find a clear proof which we address in post.

Note that even though the volume and sine-of-side gives a nice analogy with the planar case, we lose the ability to add or subtract the sides and area which is very useful in the planar case. However, this does not seem to be a serious limitation.

While the proof given in Wiki article uses 'subtracting two areas', we can achieve the result even without that. Using the notations used in the Wiki article (but assuming a spherical triangle) and denoting the volume of a tetrahedron formed by points $X$, $Y$, $Z$ and the origin by $|XYZ|$,

$\displaystyle \frac{\sin OC}{\sin OF}=\frac{|OCB|}{|OBF|}$ and $\displaystyle \frac{\sin FB}{\sin AB}=\frac{|OFB|}{|OAB|}$

Combining the two, we have $\displaystyle \frac{\sin OC}{\sin OF}=\frac{|OBC|}{|OAB|}\frac{\sin AB}{\sin FB}$

We can use a similar argument to show $\displaystyle \frac{\sin OC}{\sin OF}=\frac{|OCA|}{|OAB|}\frac{\sin AB}{\sin AF}$

Using the above two, we have $\displaystyle \frac{|OBC|}{|OCA|}=\frac{\sin FB}{\sin AF}$

We can replicate this to the other sides as well. Then,

$\displaystyle \frac{\sin AF}{\sin FB} \cdot \frac{\sin BD}{\sin DC} \cdot \frac{\sin CE}{\sin EA}=\frac{|OCA|}{|OBC|} \cdot \frac{|OAB|}{|OCA|}\cdot \frac{|OBC|}{|OAB|}=1$

The existence of Ceva's theorem can then be used to show the concurrency of medians, altitudes and angle bisectors of spherical triangles (strictly speaking, we need the converse of Ceva's, but at this point I'm taking it for granted).

Hope you enjoyed this post.

Yours Aye

Me