Mid-point of AB segment 
Let X,Y be the end points of the diameter of a circumference
  $\mathit{C}$, and let N be the mid-point of one of the arcs XY of
  $\mathit{C}$. Let A,B be two points in the segment XY. The lines NA
  and NB cut $\mathit{C}$ in the points C and D, respectively. The
  tangents to $\mathit{C}$ in C and D intersect at P. Let M the point of
  intersection between the segments XY and NP. Prove that M is the
  mid-point of the segment AB.

I found this exercise on a geometry book (olympiad book without theory, just exercises. I'm learning this type of geometry for the first time) but I do not have any idea on how to tackle it. What theorems could I use to solve this? Any help/hints will be very appreciated.
 A: 
(Please refer to the diagram.) First, we shall prove that $ABDC$ is cyclic. Let $O$ be the center of the original circle. Then, $NO\perp XY$. Therefore, $$NA.NC=NA.AC+NA^2=XA.YA+NO^2+OA^2=(XO-AO)(YO+AO)+NO^2+OA^2=XO^2+NO^2$$ We do the same thing for B and obtain $NB.ND=NA.NC$. Therefore, $ADBC$ is cyclic as claimed.
Let $\angle ANM=\theta_1$, $\angle BNM=\theta_2$, $\angle NAM=\theta_3$, $\angle NBM=\theta_4$. By cyclicity of $ABDC$, $\angle CDN=\theta_3$. As PC is tangent to the circle, $ext. \angle PCN=\theta_3$. Similarly for the angles marked $\theta_4$.
Applying sine rule to $\triangle PCN$ and $\triangle PDN$, we have, $$\frac{PC}{\sin \theta_1}=\frac{PN}{\sin \theta_3}\text {   and   }\frac{PD}{\sin \theta_2}=\frac{PN}{\sin \theta_4}\text .$$
As $PC=PD$, $$\frac{\sin \theta_1}{\sin \theta_2}=\frac{\sin \theta_3}{\sin \theta_4}\text .$$
Applying sine rule to $\triangle AMN$ and $\triangle BMN$ , we have, $$\frac{AM}{\sin \theta_1}=\frac{MN}{\sin \theta_3}\text {   and   } \frac{BM}{\sin \theta_2}=\frac{MN}{\sin \theta_4}\text .$$
$$\therefore \frac{AM}{BM}=\frac{\sin\theta_1.\sin\theta_3}{\sin\theta_2.\sin\theta_4}=1$$
$\blacksquare$

Also see:
  Power of Point, 
  Sine Rule

A: This problem can be solved in a completely straightforward don't-make-me-think way by using complex geometry (which is a frequent subject in IMO and other competitions). The most important formulas can be found HERE.
WLOG, we can assume that our circle is a unit circle in a complex plane:

Each point in the drawing is represented with a complex number. A capital letter (say $R$) is a point, the corresponding small letter ($r$) is its complex coordinate (usual convention in complex geometry).
We'll pick points $C$ and $D$ freely on the unit circle. These two points satisfy the following relation:
$$\bar{c}=\frac1c, \ \bar{d}=\frac1d\tag{1}$$
The point $A$ represents the intersection of chords $NC$ and $XY$. The point of intersection is given with the following (well-known) formula (also found in the "cheat sheet" mentioned above):
$$a=\frac{nc(x+y)-xy(n+c)}{nc-xy}$$
Notice that $n=1$, $x=i$, $y=-i$, $x+y=0$, $xy=1$. This gives:
$$a=\frac{1+c}{1-c}$$
In a similar fashion:
$$b=\frac{1+d}{1-d}$$
Denote the midpoint of AB with M:
$$m=\frac{a+b}2=\frac{1-cd}{(1-c)(1-d)}\tag{2}$$
Point P is defined by the intersection of tangents $CP$ and $CD$. Again, by a well-known formula:
$$p=\frac{2cd}{c+d}\tag{3}$$
Let us prove that points $N,M,P$ are collinear and we are done! In complex geometry, this is true iff:
$$\frac{n-m}{\bar{n}-\bar{m}}=\frac{n-p}{\bar{n}-\bar{p}}$$
Obviously $n=\bar{n}=1$. So we have to prove that:
$$\frac{1-m}{1-\bar{m}}=\frac{1-p}{1-\bar{p}}\tag{4}$$
From (2):
$$\bar{m}=\frac{1-\bar{c}\bar{d}}{(1-\bar{c})(1-\bar{d})}=\frac{1-\frac1c\frac1d}{(1-\frac1c)(1-\frac1d)}=\frac{cd-1}{(c-1))(d-1)}$$
$$\frac{1-m}{1-\bar{m}}=\frac{1-\frac{1-cd}{(1-c)(1-d)}}{1-\frac{cd-1}{(c-1))(d-1)}}=\frac{c+d-2cd}{c+d-2}\tag{5}$$
From (3):
$$\bar{p}=\frac{2\bar{c}\bar{d}}{\bar{c}+\bar{d}}=\frac{2\frac1c\frac1d}{\frac1c+\frac1d}=\frac{2}{c+d}$$
$$\frac{1-p}{1-\bar{p}}=\frac{1-\frac{2cd}{c+d}}{1-\frac{2}{c+d}}=\frac{c+d-2cd}{c+d-2}\tag{6}$$
By comparing (5) and (6) we see that (4) is true and therefore points $N,M,P$ must be collinear.
Done.
