Cho tứ giác ABCD nội tiếp đường tròn (O); ACBD=I;ADBC=S. Một đường thẳng song song với AB cắt SA,SB tại M,N. Chứng minh rằng SI là trục đẳng phương của các đường tròn ngoại tiếp các tam giác ACN,BDM.

Gọi (C1),(C2) là đường tròn ngoại tiếp các tam giác ACN,BDM. Ta có:
¯IA.¯IC=¯IB.¯IDI/(C1)=I/(C2)(1)                  
Lại có {¯SC.¯SB=¯SD.¯SA(ABCDnội tiếp)¯SN¯SB=¯SM¯SA(MN//AB)
¯SC.¯SN=¯SD.¯SMS/(C1)=S/(C2)(2)
Từ (1),(2) suy ra SI là trục đẳng phương của (C1),(C2)

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