Without loss of generality assume that one of the vertices of our triangle is the orgin

. Thus we are really examining

, where

are the other verticies. We may assume that

for some

. From here we break this into two cases, namely:

, or

. Clearly both cannot be zero since the origin and

are distinct points.

**Case 1**,

: It is clear that we assume WLOG that

and

for all other cases are analgous. Thus this triangle will have

and

for some

. Clearly seeing though that the length of

is

(since they are both on the y-axis) we may conclude that the length of

is as well. Realizing though that the lengths of

are given independently by the distance equations

and appealing once again to the equilaterallness of the triangle we may conclude that, in fact,

. Solving this equation yields

. As stated earlier though

and focusing mainly on

we see that

which is clearly irrational for any rational

.

**Case 2**,

: Now we have that

where neither

or

is zero. Constructing the midpoint

of

gives us that

and that the length of the segment

is

. Now, form

to be the median connecting

and

. If we consider the two segments

to be the restrictions of the lines

respectively we can see that

and since

we may conclue that

(now you see why we needed

). Letting

for some

again and realizing that

both lie on

we may conclude that

where

represents the change in

between the points

and

. Appealing to this we see that

. Remembering though that forming the median

bisected

we see that the new triangle formed,

is a

triangle. In particular, this tells us that

. Equating our two lengths of

gives us

and since

we may conclude that

. Thus, substitution into

gives us that

which is clearly irrational since

.

The conclusion follows.