Show that, for a circle $\bigcirc$ with given dimensions, any arc subtends the same angle at every point on $\bigcirc$.
I arrived at this problem while working on projective transformations in $\mathbb{CP}^1$. Being unfamiliar with the Inscribed Angle theorem, I was trying to generalise Thales's theorem. Posting here for verification.
The diagram below shows Thales's theorem $$\tag{1} 2\color{red}{\phi}+2\color{green}{\gamma}=\pi \implies \angle BDC=\frac{\pi}{2}.$$ 
In order to generalise Thales, consider any segment $FG \parallel BC$ on $\bigcirc$.
Since $\triangle FAD$ is isosceles $$ \tag{2} 2(\color{red}{\phi} + \color{purple}{\delta})+2\color{green}{\gamma} - \color{purple}{\psi}=\pi.$$ Appealing to $(1)$ $$\tag{3} 2\color{purple}{\delta} =\color{purple}{\psi}.$$ The same argument, applied to $\triangle ADG$ shows $$\tag{4} \angle CDG = \color{purple}\delta.$$
In general, any arc $FG$ on $\bigcirc$ subtends the same angle $\theta$, where $$\tag{5}\theta = \frac{\pi}{2} \pm\color{purple}\psi$$ depending on whether $D$ lies above or below the segment $FG$ in the above diagram.
Note: Some care has to be taken when considering points $D$ between $BC$ and $FG$, as in the following:
In this case, inspection of $\triangle FAD$ shows $(3)$ still holds and inspection of $\triangle ADG$ shows $$\tag{6} 2(\color{red}{\phi}+\color{purple}\delta+\angle FDG ) +\color{purple}\psi-2\color{red}{\phi}=\pi$$ from which the result follows.