Question

已知:如图,AB是⊙O的直径,C是⊙O上一点,∠ABC=2∠BAC,∠ABC的平分线交⊙O于点D

Answer 1

证明：∵AB是圆心O的直径

∴∠ACB=∠ADB=90°

∵∠ABC＋∠ACB＝90°,∠ABC＝2∠BAC

∴∠CAB＝30°,∠ABC＝60°

∵BD平分∠ABC

∴∠ABD＝∠CBD＝½∠ABC＝30°＝∠BAC

∴AD＝DC＝CB

⒉∵圆O的半径为2cm

∴直径AB＝4cm

∵∠ACB＝90°,∠CAB＝30°

∴CB＝½AB＝2cm

∴四边形ABCD＝AB+BC+CD+DA=AB+3BC=4＋3×2=10﹙cm﹚