3. A point in a triangle is connected to the vertices by three segments. What is the maximum number of these segments that can equal the opposite side?

#
# Answer: One.

## Solution:

Let $B M = A C$ and $A B = M C$ (see fig.). Triangles $A B M$ and $M C A$ are equal by three sides. Therefore, angle $B A M$ is equal to angle $A M C$, which means $A B \parallel M C$.

Similarly, $A C \parallel M B$.

Thus, $A B M C$ is a parallelogram, but this is not the case, because angle $B M C$, lying in this parallelogram, is greater than a straight angle.

![](https://cdn.mathpix.com/cropped/2024_05_06_507ced5c03720ddcb08bg-2.jpg?height=494&width=423&top_left_y=2060&top_left_x=1482)

## Comment on the interpretation of the condition:

The words "Point in the triangle" imply "inside the triangle." Indeed, the solution remains valid even if point $M$ is on the boundary of the triangle. If point $M$ is chosen at a vertex, we consider this to not meet the condition of the problem ("A point in the triangle is connected to the vertices by three segments" - if the point coincides with a vertex, the segment becomes degenerate).

In other words, by default, it is assumed that the point is chosen strictly inside the triangle.