如图,AB为⊙O的直径,CD为弦,且CD⊥AB,垂足为H. (1)若∠BAC=30°,求证:CD平分OB.(2)若点E为
如图,AB为⊙O的直径,CD为弦,且CD⊥AB,垂足为H.(1)若∠BAC=30°,求证:CD平分OB.(2)若点E为的中点,连接0E,CE.求证:CE平分∠OCD.(3...
如图,AB为⊙O的直径,CD为弦,且CD⊥AB,垂足为H. (1)若∠BAC=30°,求证:CD平分OB.(2)若点E为 的中点,连接0E,CE.求证:CE平分∠OCD.(3)若⊙O的半径为4,∠BAC=30°,则圆周上到直线AC距离为3的点有多少个?请说明理由.
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(1)证明见解析;(2)证明见解析;(3)2,理由见解析. |
试题分析:(1)根据圆周角定理由AB为⊙O的直径得到∠ACB=90°,而∠BAC=30°,所以∠B=60°,于是可判断△OBC为等边三角形,根据等边三角形的性质由CD⊥OB易得CD平分OB; (2)由点E为 ![](https://iknow-pic.cdn.bcebos.com/dc54564e9258d109959ce000d258ccbf6c814d00?x-bce-process=image%2Fresize%2Cm_lfit%2Cw_600%2Ch_800%2Climit_1%2Fquality%2Cq_85%2Fformat%2Cf_auto) 的中点,根据垂径定理的推论得OE⊥AB,则OE∥CD,根据平行线的性质得∠OEC=∠ECD,而∠OEC=∠OCE,所以∠OCE=∠ECD; (3)作OF⊥AC于F,交⊙O于G,根据含30度的直角三角形三边的关系得OF= ![](https://iknow-pic.cdn.bcebos.com/9358d109b3de9c82f91ab03d6f81800a19d84300?x-bce-process=image%2Fresize%2Cm_lfit%2Cw_600%2Ch_800%2Climit_1%2Fquality%2Cq_85%2Fformat%2Cf_auto) OA=2,则GF=OG-OF=2,于是可得到在弧AC上没有一个点到AC的距离为3cm,在弧AEC上有两个点到AC的距离为3cm. 试题解析:(1)证明:∵AB为⊙O的直径, ∴∠ACB=90°, ∵∠BAC=30°, ∴∠B=60°, 而OC=OB, ∴△OBC为等边三角形, ∵CD⊥OB, ∴CD平分OB; (2)证明:∵点E为 ![](https://iknow-pic.cdn.bcebos.com/dc54564e9258d109959ce000d258ccbf6c814d00?x-bce-process=image%2Fresize%2Cm_lfit%2Cw_600%2Ch_800%2Climit_1%2Fquality%2Cq_85%2Fformat%2Cf_auto) 的中点, ∴OE⊥AB, 而CD⊥AB, ∴OE∥CD ∴∠OEC=∠ECD, ∵OC=OE, ∴∠OEC=∠OCE, ∴∠OCE=∠ECD, 即CE平分∠OCD; (3)圆周上到直线AC距离为3的点有2个.理由如下: 作OF⊥AC于F,交⊙O于G,如图, ∵OA=4,∠BAC=30°, ∴OF= ![](https://iknow-pic.cdn.bcebos.com/9358d109b3de9c82f91ab03d6f81800a19d84300?x-bce-process=image%2Fresize%2Cm_lfit%2Cw_600%2Ch_800%2Climit_1%2Fquality%2Cq_85%2Fformat%2Cf_auto) OA=2, ∴GF=OG-OF=2,即在 ![](https://iknow-pic.cdn.bcebos.com/00e93901213fb80e84cf286535d12f2eb9389462?x-bce-process=image%2Fresize%2Cm_lfit%2Cw_600%2Ch_800%2Climit_1%2Fquality%2Cq_85%2Fformat%2Cf_auto) 上到AC的最大距离为2cm, ∴在 ![](https://iknow-pic.cdn.bcebos.com/00e93901213fb80e84cf286535d12f2eb9389462?x-bce-process=image%2Fresize%2Cm_lfit%2Cw_600%2Ch_800%2Climit_1%2Fquality%2Cq_85%2Fformat%2Cf_auto) 上没有一个点到AC的距离为3cm, 而在 ![](https://iknow-pic.cdn.bcebos.com/b03533fa828ba61e9a91819d4234970a304e5900?x-bce-process=image%2Fresize%2Cm_lfit%2Cw_600%2Ch_800%2Climit_1%2Fquality%2Cq_85%2Fformat%2Cf_auto) 上到AC的最大距离为6cm, ∴在 ![](https://iknow-pic.cdn.bcebos.com/b03533fa828ba61e9a91819d4234970a304e5900?x-bce-process=image%2Fresize%2Cm_lfit%2Cw_600%2Ch_800%2Climit_1%2Fquality%2Cq_85%2Fformat%2Cf_auto) 上有两个点到AC的距离为3cm. 考点: 圆的综合题. |
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