🔹 Concept of Path Problems in Triangle

When a person or object traverses equal paths along the sides of a triangle and returns to the starting point (or ends at a symmetric location), we apply the Path Angle Closure Formula:

$$\text{Angle at vertex}=\frac{{180}^{\circ }}{n}\text{where }n=\text{number of equal segments}$$

🔹 Concept Applied with 5 Equal Paths

Let’s break triangle $ABC$ into 5 equal segments and apply:

$$\mathrm{\angle }A=\frac{{180}^{\circ }}{5}={36}^{\circ }$$

🔹 Advanced Case: 7 and 9 Equal Paths

If a triangle is divided into 7 or 9 equal segments (symmetric routes), then the angle at the origin becomes:

• $\mathrm{\angle }A=\frac{{180}^{\circ }}{7}\approx {25.71}^{\circ }$
• $\mathrm{\angle }A=\frac{{180}^{\circ }}{9}={20}^{\circ }$

🔍 7-Segment Full Angle Allocation

In this triangle, we see angle divisions done using equal paths, and the total rotation follows:

$$7x={180}^{\circ }\Rightarrow x={25.71}^{\circ }$$

📏 Highest Level Path Geometry – 9 Segment Distribution

Now, triangle is broken into 9 equal segments. Each internal corner angle is $x={20}^{\circ }$. Beautiful path symmetry!