Intuition behind inversions of triangles in circles leading to similar triangles 
In here the inversion of triangle $qab$ produces triangle $q \tilde{a} \tilde{b}$ but I find it a bit weird that $\angle qab = \angle q \tilde{b} \tilde{a}$ and $\angle q ab = \angle q \tilde{a} \tilde{b}$, why exactly does the angles of triangle switch under inversion? is there something deeper that I may be missing?
Picture from Tristan_Needham's visual complex analysis book
 A: In a triangle, a larger angle is opposite a longer side. Loosely speaking, since inversion turns relatively "longer" segments into "shorter" ones, one should expect the relative angle sizes to switch.
More formally, if $|qa|\color{red}{\leq}|qb|$ so that $\angle a \color{green}{\geq} \angle b$, then because inversion switches the length comparison to $|q\tilde{a}|\color{red}{\geq}|q\tilde{b}|$, it should also switch the angle comparison to $\angle \tilde{a} \color{green}{\leq} \angle\tilde{b}$.

Of course, the fact that $\angle a \cong \angle \tilde{b}$ and $\angle b \cong \angle\tilde{a}$ follows from the triangles being similar via SAS Similarity, so we don't really need to think "comparatively". Even so, it's helpful to build and intuition about how inversion affects relative lengths and angles.
A: We can assume WLOG that $q$ is the origin and that the circle of inversion has radius one, in order that the inversion boils down to complex transformation:
$$z \to \tilde{z}=\frac{1}{\overline z}$$
Therefore, comparing the modules (= side lengths):
$$\left|\dfrac{\tilde{b}}{\tilde{a}}\right|=\left|\dfrac{1/\overline{b}}{1/\overline{a}}\right|=\left|\dfrac{\overline{a}}{\overline{b}}\right|=\left|\dfrac{a}{b}\right|$$
Therefore, triangles $qab$ and $q\tilde{b}\tilde{a}$ (pay attention to the order) have proportional lengths ; moreover they share a common angle. Consequently, they are (indirectly) similar (consider an homothety + a symmetry).
Therefore they have the same angles, but up to a symmetry.
