来个纯导角题

hbghlyj

ABCD共圆,以A,B,C,D为圆心过P作圆,圆A,B交于E,圆A,C交于F,圆B,D交于I,圆C,D交于J,圆ACF再交圆CDJ于Q,圆CDJ再交圆BDI于R,圆BDI再交圆ABE于S,圆ABE再交圆ACF于T,求证：QRST共圆
还差证明BCRT共圆、ASQD共圆就完了
∵BCRT共圆∴∠[RT,RS] = ∠[TC,SR]+∠[BR,BC]∴∠[RT,RS]-∠[QT,QS] = -∠[QT,QS]+∠[TC,SR]+∠[BR,BC]
∵ACQT共圆∴∠[QT,QS] = ∠[TC,SQ]+∠[AQ,AC]∴∠[RT,RS]-∠[QT,QS] =∠[TC,SR]-∠[TC,SQ]+∠[BR,BC]-∠[AQ,AC]
∵∠[TC,SR] = ∠[TC,SQ]-∠[SR,SQ]∴∠[RT,RS]-∠[QT,QS]= -∠[SR,SQ]+∠[BR,BC]-∠[AQ,AC]
∵BDRS共圆∴∠[SR,SQ]= -∠[SQ,SD]+∠[BR,BD]∴∠[RT,RS]-∠[QT,QS]=∠[SQ,SD]+∠[BR,BC]-∠[BR,BD]-∠[AQ,AC]
∵∠[BR,BC] =∠[BR,BD]-∠[BC,BD]∴∠[RT,RS]-∠[QT,QS]=∠[SQ,SD]-∠[AQ,AC]-∠[BC,BD]
∵ASQD共圆∴∠[SQ,SD] = ∠[AQ,AD]∴∠[RT,RS]-∠[QT,QS]= -∠[AQ,AC]+∠[AQ,AD]-∠[BC,BD]
∵∠[AQ,AC] = ∠[AQ,AD]-∠[AC,AD]
∴∠[RT,RS]-∠[QT,QS]= -∠[BC,BD]+∠[AC,AD]
∵ABCD共圆∴∠[BC,BD]=∠[AC,AD]∴∠[RT,RS]=∠[QT,QS]∴QRST共圆

hbghlyj

1. cyclic:[B,C,R,T],
   because (2)∠[TB,TC] = ∠[RB,RC].

2. ∠[TB,TC] = ∠[RB,RC],
   because (3)∠[TB,TC] = ∠[PC,PB], (4)∠[RB,RC] = ∠[PC,PB].

3. ∠[TB,TC] = ∠[PC,PB],
   because (5)∠[TB,TA] = ∠[PA,PB], (6)∠[CP,AT] = ∠[AP,TC].

4. ∠[RB,RC] = ∠[PC,PB],
   because (7)∠[RB,RD] = ∠[PD,PB], (8)∠[PC,PD] = ∠[RD,RC].

5. ∠[TB,TA] = ∠[PA,PB],
   because (9)∠[TB,TA] = ∠[EB,EA], (10)∠[PA,PB] = ∠[EB,EA].

6. ∠[CP,AT] = ∠[AP,TC],
   because (11)∠[PC,PA] = ∠[TA,TC].

7. ∠[RB,RD] = ∠[PD,PB],
   because (12)∠[RB,RD] = ∠[IB,ID], (13)∠[PD,PB] = ∠[IB,ID].

8. ∠[PC,PD] = ∠[RD,RC],
   because (14)∠[PC,PD] = ∠[JD,JC], (15)∠[RD,RC] = ∠[JD,JC].

9. ∠[TB,TA] = ∠[EB,EA],
   because cyclic:[A,B,E,T](hyp).

10. ∠[PA,PB] = ∠[EB,EA],
   because (16)∠[PA,PE] = ∠[EP,EA], (17)∠[EB,EP] = ∠[PE,PB].

11. ∠[PC,PA] = ∠[TA,TC],
   because (18)∠[PC,PA] = ∠[FA,FC], (19)∠[TA,TC] = ∠[FA,FC].

12. ∠[RB,RD] = ∠[IB,ID],
   because cyclic:[B,D,I,R](hyp).

13. ∠[PD,PB] = ∠[IB,ID],
   because (20)∠[PD,PI] = ∠[IP,ID], (21)∠[IB,IP] = ∠[PI,PB].

14. ∠[PC,PD] = ∠[JD,JC],
   because (22)∠[PC,PJ] = ∠[JP,JC], (23)∠[JD,JP] = ∠[PJ,PD].

15. ∠[RD,RC] = ∠[JD,JC],
   because cyclic:[D,C,J,R](hyp).

16. ∠[PA,PE] = ∠[EP,EA],
   because AP = AE(hyp).

17. ∠[EB,EP] = ∠[PE,PB],
   because BP = BE(hyp).

18. ∠[PC,PA] = ∠[FA,FC],
   because (24)∠[PC,PF] = ∠[FP,FC], (25)∠[FA,FP] = ∠[PF,PA].

19. ∠[TA,TC] = ∠[FA,FC],
   because cyclic:[A,C,F,T](hyp).

20. ∠[PD,PI] = ∠[IP,ID],
   because DP = DI(hyp).

21. ∠[IB,IP] = ∠[PI,PB],
   because BP = BI(hyp).

22. ∠[PC,PJ] = ∠[JP,JC],
   because CP = CJ(hyp).

23. ∠[JD,JP] = ∠[PJ,PD],
   because DP = DJ(hyp).

24. ∠[PC,PF] = ∠[FP,FC],
   because CP = CF(hyp).

25. ∠[FA,FP] = ∠[PF,PA],
   because AP = AF(hyp).



The Proof in LaTex format has been saved in  example.tex

hbghlyj

1. cyclic:[A,S,Q,D],
   because (2)∠[QD,QA] = ∠[SD,SA].

2. ∠[QD,QA] = ∠[SD,SA],
   because (3)∠[QD,QA] = ∠[PA,PD], (4)∠[SD,SA] = ∠[PA,PD].

3. ∠[QD,QA] = ∠[PA,PD],
   because (5)∠[QD,QC] = ∠[PC,PD], (6)∠[PA,PC] = ∠[QC,QA].

4. ∠[SD,SA] = ∠[PA,PD],
   because (7)∠[SD,SB] = ∠[PB,PD], (8)∠[PA,PB] = ∠[SB,SA].

5. ∠[QD,QC] = ∠[PC,PD],
   because (9)∠[QD,QC] = ∠[JD,JC], (10)∠[PC,PD] = ∠[JD,JC].

6. ∠[PA,PC] = ∠[QC,QA],
   because (11)∠[PA,PC] = ∠[FC,FA], (12)∠[QC,QA] = ∠[FC,FA].

7. ∠[SD,SB] = ∠[PB,PD],
   because (13)∠[SD,SB] = ∠[ID,IB], (14)∠[PB,PD] = ∠[ID,IB].

8. ∠[PA,PB] = ∠[SB,SA],
   because (15)∠[PA,PB] = ∠[EB,EA], (16)∠[SB,SA] = ∠[EB,EA].

9. ∠[QD,QC] = ∠[JD,JC],
   because cyclic:[D,C,J,Q](hyp).

10. ∠[PC,PD] = ∠[JD,JC],
   because (17)∠[PC,PJ] = ∠[JP,JC], (18)∠[JD,JP] = ∠[PJ,PD].

11. ∠[PA,PC] = ∠[FC,FA],
   because (19)∠[PA,PF] = ∠[FP,FA], (20)∠[FC,FP] = ∠[PF,PC].

12. ∠[QC,QA] = ∠[FC,FA],
   because cyclic:[A,C,F,Q](hyp).

13. ∠[SD,SB] = ∠[ID,IB],
   because cyclic:[B,D,I,S](hyp).

14. ∠[PB,PD] = ∠[ID,IB],
   because (21)∠[PB,PI] = ∠[IP,IB], (22)∠[ID,IP] = ∠[PI,PD].

15. ∠[PA,PB] = ∠[EB,EA],
   because (23)∠[PA,PE] = ∠[EP,EA], (24)∠[EB,EP] = ∠[PE,PB].

16. ∠[SB,SA] = ∠[EB,EA],
   because cyclic:[A,B,E,S](hyp).

17. ∠[PC,PJ] = ∠[JP,JC],
   because CP = CJ(hyp).

18. ∠[JD,JP] = ∠[PJ,PD],
   because DP = DJ(hyp).

19. ∠[PA,PF] = ∠[FP,FA],
   because AP = AF(hyp).

20. ∠[FC,FP] = ∠[PF,PC],
   because CP = CF(hyp).

21. ∠[PB,PI] = ∠[IP,IB],
   because BP = BI(hyp).

22. ∠[ID,IP] = ∠[PI,PD],
   because DP = DI(hyp).

23. ∠[PA,PE] = ∠[EP,EA],
   because AP = AE(hyp).

24. ∠[EB,EP] = ∠[PE,PB],
   because BP = BE(hyp).



The Proof in LaTex format has been saved in  example.tex

hbghlyj

需要把2#和3#整理一下