设圆心为O,半径为R.连接OP过P作OP的垂线交圆于A,B,则AB为最短的弦.
证明:过P任画一条弦交圆于C,D过O作CD的垂线垂足为E,在直角三角形PEO中,PO为斜边即PO>OE
(CD/2)^2+OE^2=R^2,(AB/2)^2+PO^2=R^2,OE^2=R^2-(CD/2)^2,PO^2=R^2-(AB/2)^2,因为OE^2
设圆心为O,半径为R.连接OP过P作OP的垂线交圆于A,B,则AB为最短的弦.
证明:过P任画一条弦交圆于C,D过O作CD的垂线垂足为E,在直角三角形PEO中,PO为斜边即PO>OE
(CD/2)^2+OE^2=R^2,(AB/2)^2+PO^2=R^2,OE^2=R^2-(CD/2)^2,PO^2=R^2-(AB/2)^2,因为OE^2