Question

已知,如图,矩形ABCD的对角线AC,于BD相交于O,OF⊥AC,交AB于点E

，交CB的延长线于F，求证AO的平方＝OE乘OF

Answer 1

OF⊥AC
∠F+∠ACB=90°
∠OAB+∠ACB=90°
∠F=∠OAB
OA=OB
∠OAB=∠OBA
∠F=∠OBA
∠FOB=∠FOB
△OBF∽△OBE
OB²=OE*OF
OA²=OE*OF

Answer 2

证明：矩形ABCD中，∠ABC=90度 AO=1/2*AC BO=1/2*BD 且AC=BD
∴AO=BO ∠CAB+∠OCB=90度
∴∠OAB=∠OBA
∵OF⊥AC
∴∠F+∠OCB=90度
∴∠F=∠OAB
∴∠F=∠OBA
∵∠BOE=∠FOB
∴△BOE∽△FOB
∴OE/OB=OB/OF
即OB^2=OE*OF
∵AO=BO
∴OA^2=OE*OF

Answer 3

∵四边形ABCD是矩形，∴OA=1/2AC,OB=1/2BD,AC=BD,∠ABC=90º
∴OA=OB,∴∠OAB=∠OBA,∵OF⊥AC,∴∠OAB+∠OEA=90º,∵∠F+∠FEB=90,∠FEB=∠AEO,∴∠F=∠OAB,∵∠OAB=∠OBA,∴∠F=∠OBF
∵∠FOB=∠BOE,∴ΔOBE∽ΔOFB,∴OB:OF=OE:OB,
∴OB²=OE•OF,∵OB=OA,∴OA²=OE•OF