三角形的五心及其性质是什么…( 五 ) _角形

延长射线AG,交BC于D,继续延长,使得GD = DE = AG/2
连接EB,EC,
四边形GBEC为平行四边形
EB = GC
延长射线PG,
过点B作PG的延长线的垂线,垂足为F
过点E作PG的延长线的垂线,垂足为H
BE与PG的延长线的交点为点Q
则,因GC//BE,角CGP = 角EQG = 角BQF
GH = GEcos(EGH) = GAcos(AGP)
HF = EBcos(BQF) = GCcos(EQG) = GCcos(CGP)
而
GH + HF = GF = GBcos(BGF) = GBcos(PI-BGP) = -GBcos(BGP),
因此,
GAcos(AGP) + GBcos(BGP) + GCcos(CGP) = 0,
GA^2 + GB^2 + GC^2 + 3PG^2
= PA^2 + PB^2 + PC^2 + 2PG[GAcos(AGP) + GBcos(BGP) + GCcos(CGP)]
= PA^2 + PB^2 + PC^2
利用上面的结论,
令P与A重合,有
GA^2 + GB^2 + GC^2 + 3GA^2
= AB^2 + AC^2 (1)
令P与B重合,有
GA^2 + GB^2 + GC^2 + 3GB^2
= AB^2 + BC^2 (2)
令P与C重合,有
GA^2 + GB^2 + GC^2 + 3GC^2
= BC^2 + AC^2 (3)
(1),(2),(3)相加,有
3[GA^2 + GB^2 + GC^2] + 3[GA^2 + GB^2 + GC^2] = 2[AB^2 + BC^2 + AC^2],
GA^2 + GB^2 + GC^2 = [AB^2 + BC^2 + AC^2]/3 = (a^2 + b^2 + c^2)/3
得证
【三角形的五心及其性质是什么…】