In this diagram, there is a circle centred at $A$. $GH$ is the perpendicular bisector of $CD$. Fixing the position of $C$, the path created by point $H$ when $D$ is moving on the circle is plotted in red.
What is the equation of the red curve? Is there any elegant way to obtain its equation or prove what type of curve it is?
Polar coordinates are hard to work with because we need to express the equation of perpendicular bisector. When working in certisian coordinate, I get a very complex paramatric equation, which is annoying. I only know that the $x$ coordinate of $H$ is
$$
x_H=p(c-1)+q+r(c-1)^{-1}$$
where c is the cosine of the argument of vector $OH$
A solution is tedious, but tractable. In parametric coordinates, parametrized by the angle $\theta = \angle DAB$, we have $$(x,y) = \left( \frac{2 \cos (\theta )+u^2+1}{2 (u+1)},-\frac{2 \cos (\theta )+ u^2-2u-1}{2 (u+1)} \cot \frac{\theta}{2} \right),$$ where $u \in (-1,1)$ is the $x$-coordinate of $C$. In implicit coordinates, $$(u+1) (x-1)^2 (-u+2 x+1)-y^2 \left(u^2-2 (u+1) x+3\right) = 0.$$
The method of solution amounts to placing the figure on the coordinate plane, and first computing the equation of the perpendicular bisector $GH$. This is done by computing the midpoint of $CD$, and taking the negative reciprocal slope of $CD$ as the slope of the perpendicular bisector, and using the point-slope formula for a line.
Then we take the equation of the line $BD$ and solve the linear system of equations to obtain the intersection point $H$, which after simplification, results in the parametric form of the locus described above. To get the implicit equation, you have to eliminate $\theta$ from the system describing the parametric locus. These computations are left as an exercise.