1. **Stating the problem:** We have an isosceles triangle with vertices Zero (top vertex), Alawi (bottom-left), and Perin (bottom-right). The angle at Zero between the routes to Alawi and Perin is $40^\circ$. 2. **Given:** - Triangle Zero-Alawi-Perin is isosceles with Zero as the apex. - Angle at Zero: $\angle AZP = 40^\circ$. - Bulbao is a point inside the triangle equidistant from all three vertices (Zero, Alawi, Perin). 3. **Goal:** Find the angle between the routes from Alawi and Perin to Bulbao, i.e., the angle at Bulbao formed by the lines Bulbao-Alawi and Bulbao-Perin. 4. **Key observations and formulas:** - Since Bulbao is equidistant from all three vertices, it is the circumcenter of the triangle. - The triangle is isosceles with $ZA = ZP$. - The angle at Zero is $40^\circ$, so the base angles at Alawi and Perin are equal. - The circumcenter lies on the perpendicular bisector of the base $AP$. 5. **Step 1: Find the base angles.** Sum of angles in triangle is $180^\circ$. $$\angle A + \angle P + \angle Z = 180^\circ$$ Since $\angle A = \angle P$ and $\angle Z = 40^\circ$: $$2\angle A + 40^\circ = 180^\circ$$ $$2\angle A = 140^\circ$$ $$\angle A = 70^\circ$$ 6. **Step 2: Use the circumcenter property.** The circumcenter is equidistant from all vertices and lies inside the triangle. 7. **Step 3: Find the angle at Bulbao between Alawi and Perin.** The angle at Bulbao between Alawi and Perin is the angle subtended by the segment $AP$ at the circumcenter. In a triangle, the angle subtended by a chord at the center of the circumscribed circle is twice the angle subtended at the circumference. Here, the angle at Zero (apex) is $40^\circ$, which is the angle subtended by base $AP$ at the circumference. Therefore, the angle at the circumcenter (Bulbao) subtended by $AP$ is: $$2 \times 40^\circ = 80^\circ$$ 8. **Final answer:** The angle between the routes from Alawi and Perin to Bulbao is $80^\circ$. **Answer:** $80^\circ$