Q:

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.

A:

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. 

