∵AG=AH，∴∠AGH=∠AHG， 从而∠AGB=∠AHD．
B
（第21题）
D
∴△ABG≌△ADH． ··········································································································· 8分
∴AB AD．
∵四边形ABCD是平行四边形，
∴四边形ABCD是菱形． ·················································································· 10分 22．（1）∵x，y都是正整数，且y
6
2，3，6． ，∴x 1，
x
∴P············································································· 4分 ，6)，P2(2，3)，P3(3，2)，P4(61)， ·1(1（2）从P1，P2，P3，P4中任取两点作直线为： P1P2，P1P3，P1P4，P2P3，P2P4，P3P4．
∴不同的直线共有6条． ··································································································· 9分 （3）∵只有直线P2P4，P3P4与抛物线有公共点，
∴从（2）的所有直线中任取一条直线与抛物线有公共点的概率是
k 1 2k b
23．（1）由 ，解得
3 k b b
21
···················· 12分 ·
63
4
3，所以y 4x 5 ·············································· 4分 5333
0)，D(0)． （2）C( ，
在Rt△OCD中，OD ∴tan OCD
5453
55，OC ， 34
OD4
··································································································· 8分 ．
OC3
1)， （3）取点A关于原点的对称点E(2，
则问题转化为求证 BOE 45 ． 由勾股定理可得，
OE 5，BE 5，OB ，
∵OB2 OE2 BE2， ∴△EOB是等腰直角三角形． ∴ BOE 45 ．
∴ AOB 135°． ·············································································································· 12分
24．（1）在△ABC中，∵AC 1，AB x，BC 3 x．
1 x 3 x∴ ，解得1 x 2． ······················································································ 4分
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