# How to draw line with 2d thickness using Animation Nodes?

I trying to make something like this (callout title. like in image) with animation nodes. And i need to draw line with thickness and animate it. If i use curve with Bevel Object (second curve) it give ugly result near twist (image1). I know how to draw curve or line mesh using animation nodes but they havent thickness. Does anyone know how to add thickness?

Let the thickness be $$T$$, the length of the base be $$L$$, the length of the other side be $$W$$, and the angle between the base and the other side is $$\theta$$. Then you should create a mesh composed of 6 vertices with the following vertices locations: $$(0,0)$$, $$(0,T)$$, and $$(L,0)$$ are trivial. For the point $$(L-T\cot{(\theta/2)}, T)$$, the y location is trivial, the x location is computed by noting that $$T\cot{(\theta/2)} = T\frac{\text{base}}{T} = \text{base}$$, where $$\text{base}$$ is the base of the triangle composed from $$(L,0)$$, the point, and its projection on the x axis. For the point $$(L+W\cos{(\theta)}, W\sin{(\theta)})$$, $$W\cos{(\pi - \theta)} = W\frac{\text{base}}{W} = \text{base}$$ where $$\text{base}$$ is the base of the triangle composed of $$(L,0)$$, the point, and the projection of the point on the x axis. And $$W\sin{(\pi - \theta)} = W\frac{\text{height}}{W} = \text{height}$$ where $$\text{height}$$ is the height of the triangle composed of $$(L,0)$$, the point, and the projection of the point on the x axis. Finally, for $$(L+\cos{(\pi-\theta)}(W-T), \sin{(\pi-\theta)}(W+T)))$$, we add a vector to the aforementioned point, this vector is computed from the triangle composed from $$(L-T\cot{(\theta/2)}, T)$$, its projection on the the side whose length is $$W$$, and the intersection point between the projection lines of the point on the x axis and the side length whose length is $$W$$. Which conclude the derivation of the point locations.