![](https://video.ask-data.xyz/img.php?b=https://iknow-pic.cdn.bcebos.com/29381f30e924b899870ba7896d061d950a7bf648?x-bce-process=image%2Fresize%2Cm_lfit%2Cw_600%2Ch_800%2Climit_1%2Fquality%2Cq_85%2Fformat%2Cf_auto)
解:(1)如答图1,连接OG.
∵EG为切线,∴∠KGE+∠OGA=90°,
∵CD⊥AB,∴∠AKH+∠OAG=90°,
又OA=OG,∴∠OGA=∠OAG,
∴∠KGE=∠AKH=∠GKE,
∴KE=GE.
(2)AC∥EF,理由为:
![](https://video.ask-data.xyz/img.php?b=https://iknow-pic.cdn.bcebos.com/b999a9014c086e062c9fe1fa01087bf40ad1cb48?x-bce-process=image%2Fresize%2Cm_lfit%2Cw_600%2Ch_800%2Climit_1%2Fquality%2Cq_85%2Fformat%2Cf_auto)
连接GD,如答图2所示.
∵KG
2=KD?GE,即
=
,
∴
=
,又∠KGE=∠GKE,
∴△GKD∽△EGK,
∴∠E=∠AGD,又∠C=∠AGD,
∴∠E=∠C,
∴AC∥EF;
(3)连接OG,OC,如答图3所示.
![](https://video.ask-data.xyz/img.php?b=https://iknow-pic.cdn.bcebos.com/8b13632762d0f703ae854b6d0bfa513d2697c548?x-bce-process=image%2Fresize%2Cm_lfit%2Cw_600%2Ch_800%2Climit_1%2Fquality%2Cq_85%2Fformat%2Cf_auto)
sinE=sin∠ACH=
,设AH=3t,则AC=5t,CH=4t,
∵KE=GE,AC∥EF,∴CK=AC=5t,∴HK=CK-CH=t.
在Rt△AHK中,根据勾股定理得AH
2+HK
2=AK
2,
即(3t)
2+t
2=(2
)
2,解得t=
.
设⊙O半径为r,在Rt△OCH中,OC=r,OH=r-3t,CH=4t,
由勾股定理得:OH
2+CH
2=OC
2,
即(r-3t)
2+(4t)
2=r
2,解得r=
t=
.
∵EF为切线,∴△OGF为直角三角形,
在Rt△OGF中,OG=r=
,tan∠OFG=tan∠CAH=
=
,
∴FG=
=
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