24.(9分)证明:(1)如图(一),连接AB,1CO ∵AC为⊙2O的直径 ∴DBAB⊥ ∴AD为⊙1O的直径 ∴1O在AD上 又1COAD⊥,1O为AD的中点 ∴△ACD是以AD为底边的等腰三角形 ∴ACCD= ············································································ (3分) (2)如图(二),连接1AO,并延长1AO交⊙1O与点E,连ED ∵四边形AEDB内接于⊙1O ∴ABCE∠=∠ 又∵ACAC= ∴1EAOC∠=∠ ∴1//COED 又AE为⊙1O的直径 ∴EDAD⊥ ∴1COAD⊥ ·········································································· (3分) (3)如图(三),连接1AO,并延长1AO交⊙1O与点E,连ED ∵1BEOC∠=∠ 又EB∠=∠ ∴1EOCE∠=∠ ∴1//COED 又EDAD⊥ ∴1COAD⊥ ·········································································· (3分)
@唏嘘雨季 数学最后一道题答案 我看不懂!!!! 什么叫做 【又∵ AC=AC】 靠 倒霉死了 烦死了