如图,点C是以AB为直径的圆O上一点,直线AC与过点B的切线相交于点D,D点E是BD的中点,直线CE交直线AB与点
如图,点C是以AB为直径的圆O上一点,直线AC与过点B的切线相交于点D,D点E是BD的中点,直线CE交直线AB与点.(1)求证:CF是⊙O的切线;(2)若ED=,tanF...
如图,点C是以AB为直径的圆O上一点,直线AC与过点B的切线相交于点D,D点E是BD的中点,直线CE交直线AB与点. (1)求证:CF是⊙O的切线;(2)若ED= ,tanF= ,求⊙O的半径.
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试题分析:(1)连CB、OC,根据切线的性质得∠ABD=90°,根据圆周角定理由AB是直径得到∠ACB=90°,即∠BCD=90°,则根据直角三角形斜边上的中线性质得CE=BE,所以∠BCE=∠CBE,所以∴OBC+∠CBE=∠OCB+∠BCE=90°,然后根据切线的判定定理得CF是⊙O的切线. (2)CE=BE=DE= ![](https://iknow-pic.cdn.bcebos.com/0b46f21fbe096b6384d4bae20f338744eaf8acb7?x-bce-process=image%2Fresize%2Cm_lfit%2Cw_600%2Ch_800%2Climit_1%2Fquality%2Cq_85%2Fformat%2Cf_auto) ,在Rt△BFE中,利用正切的定义得 ![](https://iknow-pic.cdn.bcebos.com/6a600c338744ebf86dbc808ddaf9d72a6159a7b7?x-bce-process=image%2Fresize%2Cm_lfit%2Cw_600%2Ch_800%2Climit_1%2Fquality%2Cq_85%2Fformat%2Cf_auto) ,可计算出BF=2,再利用勾股定理可计算出EF= ![](https://iknow-pic.cdn.bcebos.com/cb8065380cd79123010870c6ae345982b3b7809c?x-bce-process=image%2Fresize%2Cm_lfit%2Cw_600%2Ch_800%2Climit_1%2Fquality%2Cq_85%2Fformat%2Cf_auto) ,所以CF=CE+EF=4,然后在Rt△OCF中,利用正切定义可计算出OC. 试题解析:(1)如图,连接CB、OC, ∵BD为⊙O的切线,∴DB⊥AB。∴∠ABD=90°. ∵AB是直径,∴∠ACB=90°. ∴∠BCD=90°. ∵E为BD的中点,∴CE="BE." ∴∠BCE=∠CBE. 而∠OCB=∠OBC, ∴∠OBC+∠CBE=∠OCB+∠BCE=90°. ∴OC⊥CF, ∴CF是⊙O的切线; (2)解:CE=BE=DE= ![](https://iknow-pic.cdn.bcebos.com/0b46f21fbe096b6384d4bae20f338744eaf8acb7?x-bce-process=image%2Fresize%2Cm_lfit%2Cw_600%2Ch_800%2Climit_1%2Fquality%2Cq_85%2Fformat%2Cf_auto) , 在Rt△BFE中, ![](https://iknow-pic.cdn.bcebos.com/6a600c338744ebf86dbc808ddaf9d72a6159a7b7?x-bce-process=image%2Fresize%2Cm_lfit%2Cw_600%2Ch_800%2Climit_1%2Fquality%2Cq_85%2Fformat%2Cf_auto) ,∴BF=2. ∴ ![](https://iknow-pic.cdn.bcebos.com/d62a6059252dd42a56c4e519003b5bb5c8eab8b7?x-bce-process=image%2Fresize%2Cm_lfit%2Cw_600%2Ch_800%2Climit_1%2Fquality%2Cq_85%2Fformat%2Cf_auto) .∴CF=CE+EF=4. 在Rt△OCF中, ![](https://iknow-pic.cdn.bcebos.com/7dd98d1001e9390189ea698c78ec54e737d1969c?x-bce-process=image%2Fresize%2Cm_lfit%2Cw_600%2Ch_800%2Climit_1%2Fquality%2Cq_85%2Fformat%2Cf_auto) ,∴OC=3,即⊙O的半径为3. |
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