![](https://video.ask-data.xyz/img.php?b=https://iknow-pic.cdn.bcebos.com/e824b899a9014c0853bcc314097b02087af4f4f8?x-bce-process=image%2Fresize%2Cm_lfit%2Cw_600%2Ch_800%2Climit_1%2Fquality%2Cq_85%2Fformat%2Cf_auto)
解:如图,连接OE、OF,
∵由切线的性质可得OE=OF=⊙O的半径,∠OEC=∠OFC=∠C=90°,
∴OECF是正方形,
∵由△ABC的面积可知
×AC×BC=
×AC×OE+
×BC×OF,
∴OE=OF=
a=EC=CF,BF=BC-CF=0.5a,GH=2OE=a,
∵由切割线定理可得BF
2=BH?BG,
∴
a
2=BH(BH+a),
∴BH=
a或BH=
a(舍去),
∵OE∥DB,OE=OH,
∴△OEH∽△BDH,
∴
=
,
∴BH=BD,CD=BC+BD=a+
a=
a.
故答案为:
a.