证明 在EF上取点H，使∠CHE=∠CDF，那么C,H,F,D四点共圆, 则 EF*EH=ED*EC. (1)因为∠ABC+∠ADC=180°，所以∠CHF=∠EBC，即C,H,E,B四点共圆, 则 EF*FH=FB*FC. (2) (1)+(2)得: EF^2=ED*EC+FB*FC (3) 根切线定理得: PE^2=ED*EC. (4) QF^2=FB*FC. (5) (4)+(5)得: PE 展开
证明 在EF上取点H，使∠CHE=∠CDF，那么C,H,F,D四点共圆, 则 EF*EH=ED*EC. (1)因为∠ABC+∠ADC=180°，所以∠CHF=∠EBC，即C,H,E,B四点共圆, 则 EF*FH=FB*FC. (2) (1)+(2)得: EF^2=ED*EC+FB*FC (3) 根切线定理得: PE^2=ED*EC. (4) QF^2=FB*FC. (5) (4)+(5)得: PE^2+QF^2=ED*EC+FB*FC (6) 故得:PE^2+QF^2=EF^2.证毕. 收起
