如图,已知在△ABC中,∠ACB=90°,圆O的圆心O点在BC上,且AB切圆O与点D,若OC：OB=1:3,AD=2,求

呀,OB=3OC

如图: OB=3OC CC=OD=OE=r

OB=3r

勾股定理: BD^2=OB^2-OD^2

BD=√(9r^2-r^2)=2√2 r

BC=OB+OC=4r

AD=AC=2

AB=AD+BD=2√2r+2

勾股定理: AB^2=BC^2+AC^2

(2√2r+2)^2=(4r)^2+2^2

8r^2+8√2r+4=16r^2+4

8r^2-8√2r=0

r=0 (舍去) r=√2.

BE=OB-OE=3r-r=2r=2√2.

(上面所有^2表示平方)

先前看错了,SORRY.