A convex quadrilateral has diagonals that are perpendicular to each other. Consider the sum of the diameters of the 4 circles inscribed in the parts established by these diagonals. How can this sum be expressed using the diagonals and the perimeter of the quadrilateral?
Let's indulge in traveling from one vertex of the quadrilateral along the diagonal. Until we reach the point of tangency of the inscribed circle on the left, we will have traveled the same distance as if we had traveled along the side to the left from the chosen vertex to the circle, since the tangents drawn from an external point to a circle are equal. Continuing along the diagonal, the distance left to the intersection of the two diagonals is equal to the radius of the left circle, since its center, together with the points of tangency on the diagonals and the intersection of the diagonals, forms a square. Turning left at the intersection of the diagonals, there is again a radius-length distance left to the point of tangency of the left circle. To reach the vertex, we will use up the remaining part of the already started side. Turning back at the vertex and traveling again to the intersection of the diagonals, we continue our journey by turning left again at the intersection. Upon reaching the vertex, always turn back, and at the intersection of the diagonals, always turn left. When we return to the starting point, on the one hand, we will have traveled the diagonals twice, and on the other hand, this distance will be equal to the sum of the four sides and the four diameters. Therefore, the sum of the four diameters is obtained by subtracting the perimeter of the quadrilateral from twice the sum of the lengths of the two diagonals.