23．（本题8分）
（1）证明：∵四边形ABCD中，E、F、G、H分别是AD、BC、BD、AC的中点，
1111
∴FG＝CD，HE＝CD，FH＝AB，GE＝AB． ·············································· 2分
2222∵AB＝CD，
∴FG＝FH＝HE＝EG． ················································································ 3分 ∴四边形EGFH是菱形． ·············································································· 4分 （其他方法参照给分）
（2）解：∵四边形ABCD中，G、F、H分别是BD、BC、AC的中点，
∴GF∥DC，HF∥AB． ···················································································· 5分 ∴∠GFB＝∠DCB，∠HFC＝∠ABC． ∴∠HFC＋∠GFB＝∠ABC＋∠DCB＝90°．
∴∠GFH＝90°． ·························································································· 6分 ∴菱形EGFH是正方形． ················································································· 7分 11
∵AB＝1，∴EG＝AB＝．
22
11
∴正方形EGFH的面积＝()2＝． ····································································· 8分
24
24．（本题8分）
解：（1）直线BD与⊙O相切．理由如下：
连接OB．
∵CA是⊙O的直径，
∴∠ABC＝90°． ······································· 1分
∵OB＝OC，
∴∠OBC＝∠C．
又∵∠DBA＝∠C，
∴∠DBA＋∠OBA＝∠OBC＋∠OBA＝∠ABC＝90°． ············································· 2分
∴OB⊥BD．
又∵直线BD经过半径OB的外端点B， ······························································ 3分
∴直线BD与⊙O相切． ·················································································· 4分
（2）∵∠DBO＝90°，AD＝AO＝1，
∴AB＝OA＝OB＝1．
∴△AOB是等边三角形．
9
（第24题）
D A O C B ∴∠AOB＝60°． ···························································································· 5分
60π×12π
∴S扇形OBA＝＝． ················································································· 6分
3606 ∵在Rt△DBO中，BD＝DO2－BO2＝3，
113
∴S?DBO＝OB·BD＝×1×3＝． ································································ 7分
222
∴S阴影＝S? DBO－S扇形OBA＝
25．（本题8分）
解：（1）解法一：设第一个月单价降低x元，批发商销售完这批T恤获得的总利润为y
元． ······························································································································· 1分 根据题意，得y＝(70－50－x)(200＋10x)＋(40－50)×[500－(200＋10x)]
＝－10x2＋100x＋1000． ································································ 4分 批发商销售这批T恤可能亏本，理由如下：（答案不唯一，以下方法供参考）
方法一：当x＝17（或18或19）时，y＜0． ·························································· 5分 方法二：当y＝0时，x＝55＋5（负根舍去）．
又因为当55＋5＜x＜20时，y随x的增大而减小，
所以当x＝17或18或19时，y＜0． ························································ 5分 解法二：设第一个月单价降低x元，当月出售T恤获得的利润为y1元，清仓剩
余T恤获得的利润为y2元． ··············································································· 1分 根据题意，得y1＝(70－50－x)(200＋10x)＝－10x2＋4000， ······································· 3分 y2＝(40－50)×[500－(200＋10x)]＝100x－3000． ···················································· 4分 批发商销售这批T恤可能亏本，理由如下：（答案不唯一，以下方法供参考）