Question

在Rt三角形ABC中,角C=90度,以斜边AB上一点O为圆心OB为半径作圆O,圆O切AC于点E,交AB于点D

Answer 1

作DG⊥AC于G∵BC⊥AC,∴DG∥BC∴GE/EC=DO/OB=1又CE/AE=2/3∴AG/AC=1/5由DG∥BC∴∠AGD=∠ACB,∠ADG=∠ABC∴△AGD∽△ACB∴DG/CB=AG/AC=1/5BC=5DG∵⊙O切AC于E,∴OE⊥AC又BC⊥AC,DG⊥AC∴DG∥EO∥BC,又DO=BO∴EO为梯形GDBC的中位线∴EO=(GD+BC)/2=3GD又BC=5DG∴EO/BC=3/5∵EO∥BC∴∠FEO=∠FBC,∠FOE=∠FCB∴△FEO∽△FBC∴OF/CF=EO/BC=3/5