By simple angle chasing we show that . At the same time, the last condition I wrote is equivalent to quadrilateral being cyclic (because the perpendicular bisectors of three of its sides concur), which is equivalent to (angle chase again), and doing this for all three quadrilaterals we get to the conclusion.
Sa pamamagitan ng simpleng anggulo habol namin ipakita na . Kasabay nito, ang huling kalagayan ko wrote ay katumbas ng may apat na gilid na paikot (dahil ang tirik bisectors ng tatlong ng kanyang panig pagsabayin), na kung saan ay katumbas ng (anggulo habulin muli), at ginagawa ito para sa lahat ng tatlong quadrilaterals makuha namin ang sa wakas.