Japanese theorem for cyclic quadrilaterals

The Japanese theorem states that the centers of the incircles of certain triangles inside a cyclic quadrilateral are vertices of a rectangle.

Triangulate an arbitrary concyclic quadrilateral by its diagonals, this yields four overlapping triangles (each diagonal creates two triangles). The centers of the incircles of those triangles form a rectangle.

Let $\square ABCD$ be an arbitrary concyclic quadrilateral and let be $M_{1},M_{2},M_{3},M_{4}$ the incenters of the triangles $\triangle ABD,\triangle ABC,\triangle BCD,\triangle ACD$ then the quadrilateral formed by $M_{1},M_{2},M_{3},M_{4}$ is a rectangle.

Note that this theorem is easily extended to prove the Japanese theorem for cyclic polygons. To prove the quadrilateral case, simply construct the parallelogram tangent to the corners of the constructed rectangle, with sides parallel to the diagonals of the quadrilateral. The construction shows that the parallelogram is a rhombus, which is equivalent to showing that the sums of the radii of the incircles tangent to each diagonal are equal.

The quadrilateral case immediately proves the general case by induction on the set of triangulating partitions of a general polygon.

See also

• Carnot's theorem
• Sangaku
• Wasan

References

• In Search of the Japanese Theorem
• Japanese Theorem at Cut-the-Knot
• Japanese theorem, interactive proof with animation