已知AB为圆O的直径，过圆O上的点C的切线交AB的延长线于的E，AD垂直EC于点D且 交圆O于点F，连接BC，CF，AC

已知AB为圆O的直径，过圆O上的点C的切线交AB的延长线于的E，AD垂直EC于点D且 交圆O于点F，连接BC，CF，AC

1、∵AB为圆O的直径∴∠ACB=90°∵AD⊥EC∴∠ADC=90°∵CE是圆O的切线∴∠DCF=∠DAC∵F、A、B、C四点共圆∴∠DFC=∠ABC∴Rt△CDF∽Rt△ABC∴∠DCF=∠BAC∴∠BAC=∠DAC=∠FAC∴BC=CF2、∵AD=6，DE=8，∴AE=10(勾股定理）∵∠ECB＝∠EAC∴△EBC∽△ECA∴BE/EC＝BC/AC∵∠ACD＝∠ABC∴△ACD∽△ABC∴DC/BC＝AD/AC∴DC/AD＝BC/AC∴BE/EC＝DC/ADBE/EC=(DE-EC)/ADEC×(8-EC)=6BE由切割线定理：EC²=BE×AE=10BE，BE=EC²/10∴8EC-EC²=3/5EC²40EC=8EC²EC=5(EC=0舍去）∴BE=EC²/10=5²/10=2.53、延长BC与AD的延长线交于G∵∠DAC=∠BAC∠ACB=90°即AC⊥BC∴△ABG是等腰三角形∴AB=AG=AF+FG∵∠BCE=∠DCG=∠BAC=∠DCF∴∠DCG=∠DCF∵CD⊥AD(AD⊥EC)∴△FCG是等腰三角形∴CD是中线∴DF=DG=1/2FG即FG=2DF∴AB=AF+2DF