如图，五边形A，B，C，D，E既外接于圆，又内切于圆。

如图，五边形A，B，C，D，E既外接于圆，又内切于圆。
将五切点隔位相连，如此得到五个交点A3，B3，C3，D3，E3．证明：此五点共圆。

 

四边形ABCD中

由切点弦及同弦所对的圆周角

知△B3D2A∽△C3A2D

∴(B3A)/(DC3)=(D2A)/(DA2)=(AC2)/(DA2)

又∵(AC2)/(AE3)=(siC4∠AE3C2)/(siC4∠AC2E3)=(siC4∠DE3A2)/(siC4∠DA2E3)=(DA2)/(E3D)

∴(AC2)/(DA2)=(AE3)/(E3D)

∴(B3A)/(DC3)=(AE3)/(E3D)

即:(B3A)/(AE3)=(DC3)/(E3D)=a

(CB3)/(D3C)=a

(C3B)/(BA3)=a

(A3E)/(ED3)=a

(ED3)/(A3E)=a

∴(B3A)/(AE3)=(DC3)/(E3D)= (CB3)/(D3C)= (C3B)/(BA3)= (A3E)/(ED3)= (ED3)/(A3E)=a=1

∴B3A=AE3,C3B=BA3,D3C=CB3,E3D=DC3,A3E=ED3.

∵△B3D2A∽△C3A2D

∴E4B3=E4C3.

同理:A4C3=A4D3,B4D3=B4E3,C4E3=C4A3,D4A3=D4B3

在四边形AKA4B4中由引理(2)得:B3,C3,D3,E3共圆.

同理其它四组四点共圆.

∴B3,C3,D3,E3,A3共圆.

四边形ABCD中

    由切点弦及同弦所对的圆周角可知△B3D2A∽△C3A2D

∴(B3A)/(DC3)=(D2A)/(DA2)=(AC2)/(DA2)

又∵(AC2)/(AE3)=(sin∠AE3C2)/(sin∠AC2E3)=(sin∠DE3A2)/(sin∠DA2E3)=(DA2)/(E3D)

∴(AC2)/(DA2)=(AE3)/(E3D)

∴(B3A)/(DC3)=(AE3)/(E3D)

即:(B3A)/(AE3)=(DC3)/(E3D)=a

(CB3)/(D3C)=a

(C3B)/(BA3)=a

(A3E)/(ED3)=a

(ED3)/(A3E)=a

∴(B3A)/(AE3)=(DC3)/(E3D)= (CB3)/(D3C)= (C3B)/(BA3)= (A3E)/(ED3)= (ED3)/(A3E)=a=1

∴B3A=AE3,C3B=BA3,D3C=CB3,E3D=DC3,A3E=ED3

∵△B3D2A∽△C3A2D

∴E4B3=E4C3

同理:A4C3=A4D3
B4D3=B4E3
C4E3=C4A3
D4A3=D4B3

在四边形AKA4B4中由(2)得:B3,C3,D3,E3共圆

同理其它四组四点共圆.

∴B3,C3,D3,E3,A3共圆.

∠D2AB3=∠A2DC3，∠AD2B3=∠D2E2A2=∠DA2C3

所以∠D2B3A=∠A2C3D ==> ∠E4B3C3=∠E4C3B3 ==>E4B3=E4C3

同理A4C3=A4D3,B4D3=B4E3,C4E3=C4A3,D4A3=D4B3

△AB3D2∽△DC3A2 ==> AB3/AD2=DC3/DA2=DC3/DB2

△DC3B2∽△CD3E2 ==> DC3/DB2=CD3/CE2=CD3/CA2

△CD3A2∽△AE3C2 ==> CD3/CA2=AE3/AC2=AE3/AD2

所以AB3/AD2=AE3/AD2  ==> AB3=AE3

同理BC3=BA3，CD3=CB3，DE3=DC3，EA3=ED3

其他看楼上

表示第五排没看懂

楼上的真的很强 写这么多字~~佩服佩服楼主你就把分给他就是了 如果可以的话 加我一个