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The figure ABCD is a cyclic quadrilateral,AD and BC are produced to meet at E,AB and DC are produced to meet at F,The angle bisector of angleCED meets CD and AB at O and P respectively. SHOW THAT FP=FO
最佳解答:
關鍵在性質找出來就ok了 改天證 2012-05-12 15:29:23 補充： 圖片參考：http://imgcld.yimg.com/8/n/AD08340002/o/101205110843313869451530.jpg ∠OBE+∠OEB=∠AOP ∠OBE=∠AOP-∠OEB 弧ADC=2∠OBE 弧ABC=2∠ADC 弧ADC+弧ABC=360 2∠OBC+2∠ADC=360 ∠OBC+∠ADC=180 ∠OBC ＆ ∠ADC ---------->Supplementary angles ∠PDE=∠OBE=∠AOP-∠OEB ∠PDE+∠PED=∠DPO ∠AOP-∠OEB+∠PED=∠DPO ∴∠OEB=∠PED ∵∠AOP =∠DPO 因此 FP=FO 2012-05-12 15:32:54 補充： 再幾天後我會移除答案 請版主等我移除後再選
其他解答:
等腰三角形兩底角相等，so at Triangle OFP， if you Could prove the angle FPO (ie. ∠APO) equal angle FOP (ie. ∠DOP) => FP=FO pf： 1) ABCD is a cyclic quadrilateral，so ∠A + ∠C = 180 ° exterior angle of ∠A , that is ∠PAE = ∠C = 180 ° - ∠A-----(1) 2)The angle bisector of angle CED meets CD and AB at O and P respectively => ∠AEP = ∠OEC --------(2) 3) see Triangle PAE, ∠APO = ∠PAE + ∠AEP ,and see Triangle OEC, ∠DOP = ∠C + ∠OEC (外角=兩內對角和) 4) by (1) & (2) ，angle FPO (ie. ∠APO) = ∠PAE + ∠AEP= ∠C + ∠OEC = angle FOP (ie. ∠DOP) 5) ∠FPO = ∠ FOP => FP=FO It is really interesting！
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