Cfrgtkky

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.

(Please refer to the diagram.) First, we shall prove that $ABDC$ is cyclic. Let $O$ be the center of the original circle. Then, $NOperp XY$. 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, $$frac{AM}{sin theta_1}=frac{MN}{sin theta_3}text { and } frac{BM}{sin theta_2}=frac{MN}{sin theta_4}text .$$
$$therefore frac{AM}{BM}=frac{sintheta_1.sintheta_3}{sintheta_2.sintheta_4}=1$$
$blacksquare$

Also see:
Power of Point,
Sine Rule

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:

$$bar{c}=frac1c, bar{d}=frac1dtag{1}$$

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):

$$bar{m}=frac{1-bar{c}bar{d}}{(1-bar{c})(1-bar{d})}=frac{1-frac1cfrac1d}{(1-frac1c)(1-frac1d)}=frac{cd-1}{(c-1))(d-1)}$$

$$frac{1-m}{1-bar{m}}=frac{1-frac{1-cd}{(1-c)(1-d)}}{1-frac{cd-1}{(c-1))(d-1)}}=frac{c+d-2cd}{c+d-2}tag{5}$$

From (3):

$$bar{p}=frac{2bar{c}bar{d}}{bar{c}+bar{d}}=frac{2frac1cfrac1d}{frac1c+frac1d}=frac{2}{c+d}$$

$$frac{1-p}{1-bar{p}}=frac{1-frac{2cd}{c+d}}{1-frac{2}{c+d}}=frac{c+d-2cd}{c+d-2}tag{6}$$

By comparing (5) and (6) we see that (4) is true and therefore points $N,M,P$ must be collinear.

Done.