Question

如图,已知圆O与圆O'相交于A、B两点,点O在圆O'上,圆O'的弦OC交AB于点D.

Answer 1

证明：（1）连接OB．

∵OA=OB,

∴∠OAB=∠OBA．

∵∠OCA=∠OBA,

∴∠OAB=∠OCA．

∵∠AOC=∠DOA,

∴△AOC∽△DOA．

∴OA /OD =OC/ OA ,

∴OA2=OC•OD．

（2）∵△AOC∽△DOA,

∴AC /OC =DA/ OA ．

同理可得,BC/ OC =DB/ OB ．

∴AC/ OC +BC /OC =DA /OA +DB/ OB ,

即(AC+BC)/ OC =AB/ OA ．

∵AC+BC=根号3OC,OA=r,

∴AB=根号3r．