![](https://video.ask-data.xyz/img.php?b=https://iknow-pic.cdn.bcebos.com/4610b912c8fcc3ce212119829145d688d53f209a?x-bce-process=image%2Fresize%2Cm_lfit%2Cw_600%2Ch_800%2Climit_1%2Fquality%2Cq_85%2Fformat%2Cf_auto)
解:(1)∵AB是⊙O的直径,CD⊥AB,BF∥CD,
∴BF⊥AB,
∵点B在圆上,
∴BF是⊙O的切线;
(2)如图1,连接BD.
∵AB是⊙O的直径,
∴∠ADB=90°(直径所对的圆周角是直角);
又∵DE⊥AB
∴AD
2=AE?AB;
∵AD=8cm,AB=10cm,
AE=6.4cm,
∴BE=AB-AE=3.6cm;
(3)连接BC.
四边形CBFD为平行四边形,则四边形ACBD是正方形.理由如下:
∵四边形CBFD为平行四边形,
∴BC∥FD,即BC∥AD;
∴∠BCD=∠ADC(两直线平行,内错角相等),
∵∠BCD=∠BAD,∠CAB=∠CDB,(同弧所对的圆周角相等),
∴∠CAB+∠BAD=∠CDB+∠ADC,即∠CAD=∠BDA;
又∵∠BDA=90°(直径所对的圆周角是直角),
∴∠CAD=∠BDA=90°,
∴CD是⊙O的直径,即点E与点O重合(或线段CD过圆心O),如图2,
![](https://video.ask-data.xyz/img.php?b=https://iknow-pic.cdn.bcebos.com/2fdda3cc7cd98d109313dc80223fb80e7aec90b8?x-bce-process=image%2Fresize%2Cm_lfit%2Cw_600%2Ch_800%2Climit_1%2Fquality%2Cq_85%2Fformat%2Cf_auto)
在△OBC和△ODA中,
∵
,
∴△OBC≌△ODA(SAS),
∴BC=DA(全等三角形的对应边相等),
∴四边形ACBD是平行四边形(对边平行且相等的四边形是平行四边形);
∵∠ACB=90°(直径所对的圆周角是直角),AC=AD,
∴四边形ACBD是正方形.