21. 如图，AB是$\odot O$的弦，$OD⊥ OB$，交AB于E，且$AD=ED$. 求证：AD是$\odot O$的切线.

21. 证明:如图,连接OA,$\because OA=OB,\therefore ∠B=∠OAB.\because AD=ED,\therefore ∠EAD=∠AED.\because OD⊥OB,\therefore ∠B+∠OEB=90°.\because ∠OEB=∠AED,\therefore ∠OAB+∠EAD=90°$,即$∠OAD=90°$,$\therefore AD$是$\odot O$的切线.

22. 如图,已知AB是$\odot O$的直径,点D在$\odot O$上,C是$\odot O$外一点.若$AD// OC$,直线BC与$\odot O$相交,判断直线CD与$\odot O$的位置关系,并说明理由.

22. 相交,理由:如图,连接OD,$\because AD// OC,\therefore ∠ADO=∠DOC,∠A=∠BOC,\because OA=OD,\therefore ∠A=∠ADO,\therefore ∠DOC=∠BOC.\because OB=OD,OC=OC$,$\therefore △DOC≌△BOC$(SAS),$\therefore ∠OBC=∠ODC$.$\because$直线BC与$\odot O$相交,$\therefore ∠ABC≠90°$,$\therefore ∠ODC≠90°$,$\therefore$直线CD与$\odot O$相交.$\therefore$直线CD与$\odot O$的位置关系是相交.