设最左边的圆与AP切于E点,因DA=DP,所以DE垂直AC,DE^2=DP^2-PE^2, DE^2=DC^2-EC^2
即DP^2-PE^2=DC^2-EC^2, 即DC^2-DP^2=EC^2-PE^2, 即(DC+DP)(DC-DP)=(EC+PE)(EC-PE)
因左:DC+DP=DC+DA=3AC,DC-DP=AB-BC=AP-PC(三角形每一顶点到两切点的长相等),
右:EC+PE=EC+AE=AC,EC-PE=PC, 代入后得:3AC*(AP-PC)=AC*PC, 得3AP=4PC
由1)R1/R2=AP/PC=4/3
即DP^2-PE^2=DC^2-EC^2, 即DC^2-DP^2=EC^2-PE^2, 即(DC+DP)(DC-DP)=(EC+PE)(EC-PE)
因左:DC+DP=DC+DA=3AC,DC-DP=AB-BC=AP-PC(三角形每一顶点到两切点的长相等),
右:EC+PE=EC+AE=AC,EC-PE=PC, 代入后得:3AC*(AP-PC)=AC*PC, 得3AP=4PC
由1)R1/R2=AP/PC=4/3