# Intermediate Value Theorem in geometry - in angles?

I came across a maths problem, which I reduced to need the following result:

If $$A,B$$ are fixed points, $$C,D$$ are points not on line AB, and CD is a curve (in this case, a section of the circumference of the circle), and $$\angle ACB>\angle ADB,$$ is it necessarily true that $$\forall \alpha$$ such that$$\angle ACB<\alpha<\angle ADB$$, $$\exists$$ a point $$P$$ on the curve such that $$\angle APB=\alpha?$$

I searched up on the internet with no result. Since the angle value is 'continous', I feel like this should be true, but I can't prove it (except the easy case where $$CD$$ is a line). Any idea on how to prove this in general (when $$CD$$ is a curve)?

Any help appreciated.

When $$c :[0,l]\rightarrow \mathbb{E}^2$$ is a unit speed path with $$c(0)=C$$ and $$c(l)=D$$, then $$m=|A-c(t)|,\ n=|B-c(t)|$$ so that $$f(t) = \frac{m^2+n^2-|A-B|^2}{2mn} =\cos\ \alpha(t)$$
Since $$m,\ n$$ are continuous functions, then so is $$f$$. When $$f(t)\in [0,\pi)$$ (since the curve is not in $$[AB]$$), then note that $${\rm arccos}$$ is injective and continuous on $$[0,\pi)$$. Hence $$\alpha ={\rm arccos}\ f(t)$$ is continuous.
You need to assume the plane curve $$CD$$ does not intersect the line $$AB$$. Then indeed $$P\mapsto\angle APB$$ is continuous function of the point $$P$$ on curve $$CD$$.
Sketch: WLOG let $$A=-1, B=1$$ in $$\mathbb{C}$$, and $$(CD)\subseteq\mathbb{H}$$. Then we have a choice of $$\arg\colon\mathbb{H}\mapsto(0,\pi)$$. So we can define the oriented angle $$\angle APB$$ as $$\arg(z-1)-\arg(z+1)$$, and the unoriented angle by dropping the sign, etc.