贴下我的解法:
易得∠FPD=∠DPC=pi/4,∠FDP=∠AFP=∠DCP(FPCB四点共圆)
所以△AFP∽△ADF,△FDP∽△PDC,
AF/AD=FP/FD=PD/DC => PD*AD=AF*DC, 有AP*AD=AF^2,所以AD^2=AF^2+AF*DC=AF(AF+DC), c^2+(p-b)^2=(p-a)(p-a+p-c)=>a:b:c=4:5:3, cos∠BCA=4/5
或者
tan∠PCB=2r/(a-2r)=(a+c-b)/2(b-c), tan∠ADB=tan(pi/4+∠PCB)=(a+b-c)/(3b-3c-a) = AB/BD = 2c/(a+c-b) => a:b:c=4:5:3
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