Question

rt△abc中,∠c=90°,O为AB上一点,点O为圆心,OB长为半径的圆,交BC边于点D,于AC边相切于点E.

Answer 1

作OF⊥BC于F，连接OE。则BF=DF（垂径定理）∵CD∶BD=1∶2∴CD=DF=BF∴BF∶CF=1∶2∵∠OFB=∠ACB=90°∴OF//AC∴OB/OA=BF/CF=1/2即OB=1/2OA∵OE=OB∴OE=1/2OA∵AC是⊙O的切线∴∠AEO=90°=∠ACB∴OE//BC∴AE/AC=OA/AB=2/3AE=2/3AC=2√3则OA=4，OE=2∵∠OEC=∠ECF=∠OFC=90°∴四边形OECF是矩形∴CF=OE=2则CD=DF=1