Calculate the 3rd point of an equilateral triangle from two points at any angle, pointing the "correct" way for a Koch Snowflake

Perhaps the question title needs some work.

For context this is for the purpose of a Koch Snowflake (using C-like math syntax in a formula node in LabVIEW), thus why the triangle must be the correct way. (As given 2 points an equilateral triangle may be in one of two directions.)

To briefly go over the algorithm: I have an array of 4 predefined coordinates initially forming a triangle, the first "generation" of the fractal. To generate the next iteration, one must for each line (pair of coordinates) get the 1/3rd and 2/3rd midpoints to be the base of a new triangle on that face, and then calculate the position of the 3rd point of the new triangle (the subject of this question). Do this for all current sides, concatenating the resulting arrays into a new array that forms the next generation of the snowflake.

The array of coordinates is in a clockwise order, e.g. each vertex travelling clockwise around the shape corresponds to the next item in the array, something like this for the 2nd generation:

This means that when going to add a triangle to a face, e.g. between, in that image, the vertices labelled 0 and 1, you first get the midpoints which I'll call "c" and "d", you can just rotate "d" anti-clockwise around "c" by 60 degrees to find where the new triangle top point will be (labelled e).

I believe this should hold (e.g. 60 degrees anticlockwise rotating the later point around the earlier) for anywhere around the snowflake, however currently my maths only seems to work in the case where the initial triangle has a vertical side: [(0,0), (0,1)]. Else wise the triangle goes off in some other direction.

I believe I have correctly constructed my loops such that the triangle generating VI (virtual instrument, effectively a "function" in written languages) will work on each line segment sequentially, but my actual calculation isn't working and I am at a loss as to how to get it in the right direction. Below is my current maths for calculating the triangle points from a single line segment, where a and b are the original vertices of the segment, c and d form new triangle base that are in-line with the original line, and e is the part that sticks out. I don't want to call it "top" as for a triangle formed from a segment going from upper-right to lower-left, the "top" will stick down.

cx = ax + (bx - ax)/3;
dx = ax + 2*(bx - ax)/3;


cy = ay + (by - ay)/3;
dy = ay + 2*(by - ay)/3;



dX = dx - cx;
dY = dy - cy;



ex = (cos(1.0471975512) * dX + sin(1.0471975512) * dY) + cx;
ey = (sin(1.0471975512) * dX + cos(1.0471975512) * dY) + cy;

note 1.0471975512 is just 60 degrees in radians.

Currently for generation 2 it makes this: (note the seemingly separated triangle to the left is formed by the 2 triangles on the top and bottom having their e vertices meet in the middle and is not actually an independent triangle.)

I suspect the necessity for having slightly different equations depending on weather ax or bx is larger etc, perhaps something to do with how the periodicity of sin/cos may need to be accounted for (something about quadrants in spherical coordinates?), as it looks like the misplaced triangles are at 60 degrees, just that the angle is between the wrong lines. However this is a guess and I'm just not able to imagine how to do this programmatically let alone on paper.

Thankfully the maths formula node allows for if and else statements which would allow for this to be implemented if it's the case but as said I am not awfully familiar with adjusting for what I'll naively call the "quadrants thing", and am unsure how to know which quadrant one is in for each case.

This was a long and rambling question which inevitably tempts nonsense so if you've any clarifying questions please comment and I'll try to fix anything/everything.

Answering my own question thanks to @JohanC, Unsurprisingly this was a case of making many tiny adjustments and giving up just before getting it right.

The correct formula was this:

ex = (cos(1.0471975512) * dX + sin(1.0471975512) * dY) + cx;
ey = (-sin(1.0471975512) * dX + cos(1.0471975512) * dY) + cy;

just adding a minus to the second sine function. Note that if one were travelling anticlockwise then one would want to rotate points clockwise, so you instead have the 1st sine function negated and the second one positive.