Geometry question from IMO 2009 (IMO 2009/2). Let $ABC$ be a triangle with circumcenter $O$. The points $P$
and $Q$ are interior points of the sides $CA$ and $AB$, respectively. Let $K$, $L$, and $M$ be the
midpoints of the segments $BP$, $CQ$, and $PQ$, respectively, and let $\Gamma$ be the circle passing
through $K$, $L$, and $M$. Suppose that the line $PQ$ is tangent to the circle $\Gamma$. Prove that
$OP = OQ$.
So well it is pretty easily solvable using power of a point, but the same question is also given as a exercise in the chapter of complex numbers. And m unable to solve it using complex numbers. The hints provided in the book regarding the question are as follows:
1.Find some synthetic observations first. Parallel lines. (That obvious as we have midpoints and infact to get the synthetic solution using power of a point this observation plays an important role.)
2.The condition is equivalent to the quadrilateral formed by lines KL, PQ, AB, AC
being cyclic. (Huhhhhh whattttt?! this one to me comes out of nowhere and infact m still unable to see the reasoning behind this ( i mean where does this come from and what motivates us to try that. Could somebody explain that as well?))
3.This is equivalent to $\frac{a−b}{p−q} : \frac{k−l}{a−c}∈ R$. Use Lemma 6.30 [A point $P$ lies on a chord $AB$ of the unit circle if and only if $p + ab\bar p = a + b$ and expand. ( And nope i dont understand this one either. Isnt the criterion for concyclic complex numbers slightly different in the form?)
Could someone please explain these along with a solution? M pretty confused
 A: Not sure about following the specific hints besides 1, which it sounds like you figured out: $ML || AC$ and $MK || AB$ based on midpoints.
By power of a point, we have $R^2 - PO^2 = AP\times PC$ and similarly for $Q$, which implies the condition is equivalent to showing $AP\times PC = AQ\times QB$
Now going into complex numbers, in order to simplify the math set $m = 1$ and put the origin in the center of $\Gamma$. That means $\Gamma$ is the unit circle and $|k| = |l| = 1$. Since $PQ$ is tangent and $M$ is its midpoint, $q = \bar{p}$
The parallel line observation translates to
$$
x = (a - q) / (m - k) \in \mathbb{R}
$$
$$
y = (a - p) / (m - l) \in \mathbb{R}
$$
$$
\implies a = x(1-k) +q = y(1-l) + p
$$
Conjugate both sides and add:
$$
x(2-k-\bar{k}) = y(2-l-\bar{l})
$$
$$
x(1-k)(1-\bar{k}) = y(1-l)(1-\bar{l})
$$
$$
x|(1-k)|^2 = y|(1-l)|^2
$$
Two more observations:
$$
p - k = k - b \implies b = 2k - p
$$
$$
q - l = l - c \implies c = 2l - q
$$
We can now show $AP\times PC = AQ\times QB$ directly
$$
|a-p||p-c| = |y(1-l)||p+q-2l| = 2y|1-l|^2
$$
$$
|a-q||q-b| = |x(1-k)||p+q-2k| = 2x|1-k|^2
$$
