In the convex quadrilateral $ABCD$, the intersection points of the opposite sides are $K$ and $L$, and the intersection point of the diagonals is $M$. The diagonal $AC$ intersects the line $KL$ at point $N$. Determine the length of $CN$ if it is known that $AM=3 \text{~cm}$ and $MC=2 \text{~cm}$.
We will use the following statement, known as Menelaus' Theorem:

(1) The points \(X, Y, Z\) on the sides \(AB, BC, CA\) of triangle \(ABC\) (different from the vertices) are collinear if and only if for the directed segments, the following holds:

\[
\frac{AX}{XB} \cdot \frac{BY}{YC} \cdot \frac{CZ}{ZA} = -1
\]

(A proof of this can be found, for example, in the 1260th problem of the Collection of Geometric Problems I.)

![](https://cdn.mathpix.com/cropped/2024_05_02_f46ce126f69e5da61968g-1.jpg?height=762&width=1195&top_left_y=492&top_left_x=454)

We apply (1) first to the triangle \(BMA\) and the points \(K, C, D\), then to the triangle \(BCA\) and the points \(K, N, L\), and finally to the triangle \(BCM\) and the points \(L, A, D\):

\[
\begin{aligned}
& \frac{BD}{DM} \cdot \frac{MC}{CA} \cdot \frac{AK}{KB} = -1 \\
& \frac{BK}{KA} \cdot \frac{AN}{NC} \cdot \frac{CL}{LB} = -1 \\
& \frac{BL}{LC} \cdot \frac{CA}{AM} \cdot \frac{MD}{DB} = -1
\end{aligned}
\]

Multiplying these equations and performing the possible simplifications, we get:

\[
\frac{AN}{NC} \cdot \frac{MC}{AM} = -1, \quad \text{that is} \quad \frac{CN}{AN} = \frac{MC}{AM} = \frac{2}{3}
\]

From this:

\(3CN = 2AN = 2(AC + CN)\), hence \(CN = 2AC = 2(AM + MC) = 10 \text{ cm}\).

Judit Nagy (Miskolc, Földes F. Gymnasium, 1st year)