I can't really see what else it could be. Equidistant means Equal Distance. If the points on the triangle are to be of equal distance to each other, then it's an equilateral triangle.

PandaExplosion, you'll probably need to revisit your high school Trigonometry for this. This is how I would go about it:

`Please login to see this picture.`

Look at the green triangle. You want to find out the **x **and **y **lengths, as that will basically tell you how many X and Y pixels away from **A **you must position **C**.

Trigonometry can be used to decipher right-angle triangles; if you know at least 1 acute angle and 1 side length, you can work out the other side lengths. So you split your equilateral triangle into 2, which will create 2 right-angle triangles, including the the pink one in the picture

You already know all angles of this pink triangle: 90°, 60°, and 30° (60° divided by 2). You can also find out the side length **d **from Fusion (distance between two points, divided by 2.0)

Next, find out the length of the side **e**, which is called the hypotenuse. You can find the equation to do this with google or chatGPT. If you use the 30° angle, then I believe the equation is hypotenuse = opposite/sin(30). In other words, **e = d/sin(30)**

You now have one length of the green triangle (**e**). Now you need to get an angle. Let's get **g°**, as shown in the bottom picture.

First, use Fusion to measure the angle from **A** to **B**. This will give you the angle from the **horizontal**. In other words, it will give you **f°**.

You know that the angle of your equilateral triangle at **A** is 60°. So 60-**f°** will give you **h°**

**g°** and **h°** together form 90°. So **g° = 90-h°**

You now know one side length of the green triangle (the hypotenuse **e**) and one acute angle (**g°**). You can now use the appropriate trigonometric functions to work out the other two sides: the "opposite" **x** and the "adjacent" **y. **If I'm not mistaken, it's **x = e * sin(g°)** and **y = e * cos(g°).**

I'm no maths expert, but as far as I know this would be the way to do it. You may need to tweak some things for situations where C is to the left of or above A and/or B.