Question

如圖，正方形ABCD的邊長(zhǎng)為6，點(diǎn)O是對(duì)角線AC、BD的交點(diǎn).點(diǎn)E在CD上，且DE=2CE，連接BE.過(guò)點(diǎn)C作CF⊥BE，垂足是F，連接OF，則OF的長(zhǎng)為         .

Answer 1

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試題分析：∵正方形ABCD的邊長(zhǎng)為6，DE=2CE，∴CE="2," ED=4,BD=,OB=.
∴.
∵∠BOC=∠BFC=90°,∴點(diǎn)O,B,C,F在以BC為直徑的圓上.∴∠OFB=∠OCB=45°.
∴∠OFB=∠EDB=45°.
又∵∠OBF=∠EBD,∴△OBF∽△EBD.
∴,即.