1. **Problem statement:** We have a triangle with three squares constructed externally on each of its sides. We need to find the sum of the three marked angles, each formed at the vertex where a square meets the triangle. 2. **Key idea:** Each square has right angles ($90^\circ$). The marked angles are external angles adjacent to the triangle's vertices, formed by the sides of the squares. 3. **Step-by-step reasoning:** - Let the triangle have vertices $A$, $B$, and $C$ with interior angles $\angle A$, $\angle B$, and $\angle C$. - Each square is constructed on a side of the triangle, so the square's side coincides with a triangle side, and the square extends outward. - At each vertex, the marked angle is the angle between the side of the square and the adjacent side of the triangle, outside the triangle. - Since the square has right angles, the angle between the side of the square and the adjacent side of the triangle is $90^\circ$ minus the interior angle at that vertex. - Therefore, the marked angle at vertex $A$ is $90^\circ - \angle A$, similarly for $B$ and $C$. 4. **Sum of the marked angles:** $$\text{Sum} = (90^\circ - \angle A) + (90^\circ - \angle B) + (90^\circ - \angle C)$$ 5. **Simplify:** $$= 90^\circ + 90^\circ + 90^\circ - (\angle A + \angle B + \angle C)$$ $$= 270^\circ - 180^\circ$$ $$= 90^\circ$$ 6. **Conclusion:** The sum of the three marked angles is $90^\circ$. **Final answer:** $$\boxed{90^\circ}$$