求教高中几何1-业主生活–手机房天下问答

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lovelytomorrow 丨Lv 3

已知 ABCD是圆内接四边形,AB与DC交于E，AD与BC交于，PE、QF是切线，P,Q是切点。求证 PE^2+QF^2=EF^2.证明在EF上取点H，使∠CHF=∠CDF，那么C,H,F,D四点共圆,则 EF*EH=ED*EC. (1-1))因为∠ABC+∠ADC=180°，所以∠CHE=∠EBC，即C,H,E,B四点共圆,则 EF*FH=FB*FC. (1-2)(1-1)+(1-2)得:EF^2=ED*EC+FB*FC (2)根切线定理得:PE^2=ED*EC. (3-1)QF^2=FB*FC. (3-2)(3-1)+(3-2)得:PE^2+QF^2=ED*EC+FB*FC (4)故得:PE^2+QF^2=EF^2.证完.

+1 2008-07-08 举报