It is proved that the sides of a complete quadrilateral intersect into four triangles, and their inner and outer centers are *** 16 points. In each triangle, draw the connecting lines of the inner and outer centers in pairs, so that a * * * can get 24 circles. These 24 circles intersect the inner and outer centers of each triangle, and they intersect the other 63 points. They are distributed on eight circles, and each circle has eight points. These eight circles form two groups of mutually orthogonal circles with axes of * * *, each group contains four circles, and their idempotent axes pass through Mick points of a complete quadrilateral.

It is proved that (1) makes P, PA, PB, PE, Q, QA, QD, QF, R, RB, RC, RF, S, SC, SD, SE △ABE, △ADF, △BCF, △CDE, respectively, with complete quadrangles. The intersection of PBPE, RRF, SSDPAPE, QQF, SSCPPE, QQF and RRC. The dense points P2, Q2, R2 and S2 of complete quadrilateral QFRFCDSDF, PESECBRBE, PBADSCQDE and QDABRCPBF are defined as QQF, RCRF and SCSD；; ; PBPE、RBRC、SCSE； The intersection of PPB, QQD, SSCPAPB, QQD and RRC. The Mik points P3, Q3, R3 and S3 of the complete quadrilateral QASCFDRC, PARCEBSC, PBQFDEASD and QDPEBFARB are defined as QQA, RCRF and SSE；; PPA、RRF、SCSE； PPB、QQF、SDSE； The intersection of PPE, QQD and RBRF. The dense points P4, Q4, R4 and S4 of QRCDFSC, PSCBERC, APESEDQDE and AQFRFBPBF of complete quadrilateral are defined as QQA, RRB and SCSD；; . PPA、RBRC、SSDPAPE、QAQD、SDSE； The intersection of PAPB, QAQF and RBRF.

(2) Because A, P, B, PE and A, Q, D and QF are all * * cycles, there are ∞⊙PERS =∠ Bapu =∠QFD =∞⊙QFRS, so R, S, QF and PE are all * *.

Because ∠PAR 1PE, ∠SR 1SC, ∠QAS 1QF and ∠RSRC are right angles, there are

∠PAR 1SC=∠SR 1PE，∠QAS 1RC=∠RS 1QF

∠PAR 1SC=∠PAQSC=∠SQFPE，∠QAS 1RC=∠QAPRC=∠RPEQF。

So there are R 1, S 1 ⊙O 1, that is, PE, QF, r, s, P 1, Q 1, S 1 eight-point * * cycles.

Similarly, ⊙ O2: pbqdrcscp 2q2s2; ⊙O3:pqrfsep 3q 3 r 3s 3； ⊙O4:paqarbsdp 4 q 4 r 4s 4； ⊙O5:pqarcsp 3 q 4 r 2s 1； ⊙O6:peqdrbsep 1q 2 r4s 3； ⊙O7:paqrscp 4 q3r 1s 2； ⊙O8:PBQFRFSDR2Q 1R3S4。

(3) Because ∠ p4sdpa+∠ p4scpa = 90, ∠ p4o4pa+∠ p4o7pa = 180. Therefore, P4, PA, O4 and O7 are * * cycles.

(4) Because ⊙O 1, ⊙O2, ⊙O3, ⊙O4 and ⊙O5, ⊙O6, ⊙O7 and ⊙O8 are two orthogonal groups of axis circles, there are O.

Similarly, O 1, O2 and O3 are also idempotent points of ⊙O5, ⊙O6, ⊙O7 and ⊙O8, so ⊙O 1, ⊙O2, ⊙O3 and ⊙.

(5) Let m be the Mick point of complete quadrilateral ABCDEF, and P ′, Q ′ and R ′ be the midpoint of PAPB, QAQF and RBRF. From the problem 17 of exercise 2 1 and the orthogonal relationship between ⊙O4 and ⊙O8, there are P ′, Q ′ and R ′.

∠Q′P′R′=∠Q′SDR′=∠SEC

From the review questions 1 9 and 18 of1,A, D, F, Q', M and B, C, F, R' and M are all painted with circles, Q'A=Q'F=Q'QA, R' B =

∠Q′MR′=∠FMQ′-∠FMR′=( 180-∠FAQ′)-( 180-∠FBR′)=

$ ∠ FBR ′-∠ FAQ ′ = (90-\ frac {∠ br ′ f} {2})-(90-\ frac {∠ AQ ′ f} {2}) $ =

$ \ frac {∠AQ ' F-∠BR ' F } { 2 } = \ frac {∠ADF-∠BCF } { 2 } = \ frac {∠CED } { 2 } =∠SEC $。

So ∠q'p'r'=∠q'Mr', so p', q', r'* * * circle, that is, m is on a circle with a diameter of O4O8, so ∠ O4Mo8 = 90. Similarly ∠ O 1mo.