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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.

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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.

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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.

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