From triangles BPC, CQA, ARB,

Observe that, in ΔBPC, X is the intersection of the bisectors of angles at B and C. It follows that PX is the bisector of the angle at P. Thus,

This result holds for any other pair of diagonals PX, QY, RZ.

Assume that the diagonals are not concurrent, and thus form a triangle.

Select a diagonal, RZ, say, and, through the opposite vertex v, draw rz, parallel to RZ, with z, r on QA, RB. Then, Δzvq, Δxvq are congruent (ASA). Choose y, p so that Δxvr, Δyvr are congruent (ASA) making

and ∠vpy = ∠OPY so that Δvpy, Δvpz are congruent (SAA). Then the angles of pzqxry, PZQXRY are the same.

** Informally," the mapping pzqxry ↔ abc is 1-1" suffices, but more formally the derivation of this result is as follows:

The intersections of opposite sides of pzqxry give Δabc. Since ∠zbx = β, if we draw a line at β to bz through b, bz is the bisector of the angle at b. But, rz (the bisector at r) meets bz at z, so z is the incenter of the triangle, T, say (with vertices b, r, and a third vertex at a, angle 2α with ry). Thus, its bisector az must make angle α with ry. But ra lies along ry and ∠zar = α, so the point, a, completes T. Repeating this argument, the angles of Δabc are seen to be trisected.