Hello, Bruce!

We could use a **lot** of clarification.

. . But I'll take a wild guess . . .

Consider a triangulated Triangle.

Let the vertices of all the triangles be labled $\displaystyle x,\,y,\,z$ arbitarily.

Show that the number of segments on the outside perimeter labled $\displaystyle xy$

has the same parity as the triangulation labled $\displaystyle xyz.$

First of all, I'll guess that a "triangulated triangle" is something like this

. . (not necessarily equilateral): Code:

*
/ \
* *
/ \ / \
* - * - *
/ \ / \ / \
* - * - * - *
/ \ / \ / \ / \
* - * - * - * - *

We can call this a triangulation of "order 4": $\displaystyle n = 4.$

Then each of the __smallest__ triangles is labeled with $\displaystyle x,\,y,\,z$

One arrangement looks like this: Code:

x
/ \
y z
/ \ / \
z x y
/ \ / \ / \
x - y - z - x
/ \ / \ / \ / \
y - z - x - y - z

Reading around the outside perimeter, the sequence of vertices is:

. . $\displaystyle x-y-z-x-y-z-x-y-z-x-y-z$

That is, $\displaystyle x\!-\!y\!-\!z$ occurs $\displaystyle n=4$ times.

Further sketching convinced me that this always happens.

For a triangulation of order $\displaystyle n$, the $\displaystyle x\!-\!y\!-\!z$ occurs $\displaystyle n$ times.

. . Hence, the segment $\displaystyle xy$ appears $\displaystyle n$ times.

If "the triangulation labeled $\displaystyle xyz$" means the __order__ of the triangulation,

then the number of xy-segment not only has the same parity as the order,

. . it is actually *equal* to the order.

Is this *anything* like what you meant?