Triangle Angle and Segment Intersection

Clearly the statement is satisfied with $n = 0$ (as there are no additional lines), and when $n = 1$ the line is the median of an isosceles triangle (because the median of an isosceles triangle is also the angle bisector of the angle it intersects), so we want to know if it is possible for $n \geq 2$ .

We can prove that it is impossible for $n \geq 2$ with a clever proof by contradiction. Lets start with when $n = 3$ . We are essentially wondering if it is possible for two lines to trisect the angle of a triangle and trisect the opposite side.

Triangle Diagram not drawn to scale

Let's assume $\angle ACM \cong \angle MCN \cong \angle NCB$ and $\overline{\rm AM} \cong \overline{\rm MN} \cong \overline{\rm NB}$ . Because of the segment congruency, $\overline{\rm CM}$ is the median of $\triangle ACN$ and $\overline{\rm CN}$ is the median of $\triangle MCB$ . Because the median of the triangles are also the angle bisectors of the top vertex, $\triangle ACN$ and $\triangle MCB$ are both isosceles. Since the triangles are isosceles, $\overline{\rm CM} \perp \overline{\rm AN}$ and $\overline{\rm CN} \perp \overline{\rm MB}$ (the median of an isosceles triangle is always perpendicular to the base). We can now clearly see the problem, because since $\overline{\rm CM}$ and $\overline{\rm CN}$ are both perpendicular to the same line, $\overline{\rm CM} \| \overline{\rm CN}$ , so the lines will never meet at point $C$ !

We will run into this same contradiction whenever we try to draw these lines on a triangle. The two segments next to where any of these lines intersects the base are congruent, so all the lines are medians of some isosceles triangle. Therefore the lines are all perpendicular to the base, and they will never intersect. You can only draw 0 or 1 line to evenly divide both the vertex and opposite segment of a triangle.