Let $ABC$ be a triangle, $M$ the midpoint of $BC$, and $N$ one-third of the way along $AC$ from $C$. The line $(MN)$ intersects $(AB)$ at $P$. Determine the position of $P$ relative to $A$ and $B$.
![](https://cdn.mathpix.com/cropped/2024_05_10_59677aa68548da326d08g-247.jpg?height=873&width=1046&top_left_y=1019&top_left_x=496)

The relation $\frac{M B}{M C} \cdot \frac{N C}{N A} \cdot \frac{P A}{P B}=1$ gives $\frac{P A}{P B}=2$, and $P$ is the symmetric point of $A$ with respect to $B$.

To generalize the concept of "midpoint," "third," I introduced barycenters, which require some practice with vector calculus, and demonstrated Ceva's theorem using barycentric coordinates: let three points on the sides of a triangle, $M \in (B C)$, $N \in (C A)$, $P \in (A B)$; the lines $(A M)$, $(B N)$, $(C P)$ are concurrent if and only if, in algebraic measure, $\frac{M B}{M C} \cdot \frac{N C}{N A} \cdot \frac{P A}{P B}=-1$.

Then, I recalled the Pythagorean theorem and its generalization, the Law of Cosines: $c^{2}=a^{2}+b^{2}-2 a b \cos \widehat{C}$. Along the way, I showed Heron's formula: the area of the triangle $S=\sqrt{p(p-a)(p-b)(p-c)}$.

Next, I addressed transformations of the plane: translations, rotations (composition of two rotations), homotheties, similarities, illustrating with two exercises: