The original question is as follows,
Let O be the center of a circle passing through points A, B, C and let AD be a diameter of the circle. Let the tangent line at D to the circle intersect line BC at P. Let line OP intersect line AC and AB at M and N respectively. Prove that O is the midpoint of MN.
This is what I have come up with the picture,
I have the equation of line of BC is $Z+BC\cdot\overline{\rm Z}=-BC$.
And the equation of the tangent line at D is $Z+D^{2}\overline{\rm Z}=2D$
After assuming it is a unit circle, I expressed point B and C as $e^{i\alpha}$ and $e^{i\beta}$ respectively. But I am stuck afterward.
I would appreciate if anyone can help.
Lower case letters denote the affixes of complex numbers. Let $O$ be the center of the unit circle $(ABC)$ and let line $AD$ be the real axis. So, $a=-1$ and $d=1$.
Considering the equation of line $DP$, $p$ satisfies:
$$p+\overline{p} = 2 \tag{1}$$
Since $p$ also lies on line $BC$:
$$p+bc\overline{p}=b+c \tag{2}$$
Solving $(1)$ and $(2)$ simultaneously:
$$p=\frac{-2bc+b+c}{1-bc} \tag{3}$$
The equation of the line containing $o$ (origin) and $p$ is:
$$z=\frac{2bc-b-c}{-2+c+b} \cdot \overline{z} \tag{4}$$
So, $n$ satisfies $(4)$:
$$n=\frac{2bc-b-c}{-2+c+b} \cdot \overline{n} \tag{5}$$
Since $n$ also lies on line $AB$ :
$$n=\overline{n}b-1+b \tag{6}$$
Solving $(5)$ and $(6)$ simultaneously:
$$n=\frac{-2bc+b+c}{b-c} \tag{7}$$
Likewise, $m$ satisfies $(4)$:
$$m=\frac{2bc-b-c}{-2+c+b} \cdot \overline{m} \tag{8}$$
Since $m$ also lies on line $AC$:
$$m=\overline{m}c-1+c \tag{9}$$
Solving $(8)$ and $(9)$ simultaneously:
$$m=\frac{2bc-b-c}{b-c} \tag{10}$$
Considering $(7)$ and $(10)$:
$$\fbox{$n+m=0$} \; \;\; \blacksquare$$