命题设G为△ABC的重心,AG,BG,CG与△ABC的外接圆相交于D、E、F,则AGGD+GBEG+GCFG=3.该题是《数学通报》征解题387.文[1]把它推广为:定理若P是△ABC的外接圆内的点,AP,BP,CP与外接圆交于D、E、F,O是外心,G是重心,P点落在以OG为直径的圆上的充要条件是APPD+PBEP+PCFP=3.本文把这 Proposition Let G be the center of gravity of ABC, and the circumcircle of AG, BG, CG and △ ABC intersect at D, E, F, then AGGD + GBEG + GCFG = 3. This question is solved by Math. [1] generalize it as follows: Theorem If P is the point inside the circumcircle of △ ABC, AP, BP, CP are circumscribed with circumcircle D, E, F, O are the outer center, G is the center of gravity, P is The necessary and sufficient condition for a circle with OG diameter is APPD + PBEP + PCFP = 3. This article put this