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命题 如图1,P、Q是△ABC的等角共轭点(∠PAB=∠QAC,∠PBC=∠QBA,∠PCB=∠QCA),R、S_△表示 △ABC的外接圆半径和△ABC的面积。则AP·AQ·BC+BP·BQ·AC+CP·CQ·AB=4R·S_△。
The proposition is shown in Fig. 1. P and Q are equal angle conjugate points of △ABC (∠PAB=∠QAC,∠PBC=∠QBA,∠PCB=∠QCA), and R and S_△ represent the circumcircle radius of △ABC and △. The area of ABC. Then AP·AQ·BC+BP·BQ·AC+CP·CQ·AB=4R·S_△.