>>122683741) The problem as stated gives the information that the large quadrilateral is a square. So it isn't merely a non-square rhombus.
2) From this it follows that the remainings angles of a triangle of three sectors sum to a right angle. Call one a, and all the other 90-a, say.
3) Now, since opposite sides of a square are parallel, the diagonal line set up on opposite sides of a square has the same angle on opposing sides, and pairwise supplementary angles on the other sides, which together form a straight angle on any side of the square.
4) And each of the long diagonals are congruent, since they are the hypotenuses of the large right triangles of three sectors.
5) So the quadrilateral of three sectors is a parallelogram.
6) So the long diagonals are pairwise parallel to each other.
Something something the green quadrilateral is a square because reasons, so the other angles are also right angles, so you can flip the small triangles around and that angle is also necessarily a right angle, making squares congruent to the green square, hence 1/5.
