21. (10分)如图,在单位长度为1的网格中,点$O,A,B$均在格点上,$OA=3$,$AB=2$,以$O$为圆心,$OA$为半径画圆,请按下列步骤完成作图,并回答问题:
①过点$A$作切线$AC$,且$AC=4$(点$C$在$A$的上方);
②连接$OC$,交$\odot O$于点$D$;
③连接$BD$,与$AC$交于点$E$.
(1)求证:$BD$为$\odot O$的切线;
(2)求$AE$的长度.

21.

(1)证明:$\because AC$是$\odot O$的切线,
$\therefore OA⊥AC$.
$\because OA=3,AC=4$,
$\therefore OC=\sqrt {OA^{2}+AC^{2}}=5$.
$\because OA=3,AB=2$,
$\therefore OB=OA+AB=5$,
$\therefore OB=OC$.
又$\because OD=OA=3,∠AOC=∠DOB$,
$\therefore △AOC≌△DOB(SAS)$,
$\therefore ∠OAC=∠ODB=90^{\circ },\therefore OD⊥BD$.
$\because OD$为半径,$\therefore BD$为$\odot O$的切线.
(2)解:$\because △AOC≌△DOB$,
$\therefore BD=AC=4$.
$\because ∠ABE=∠DBO,∠BAE=∠BDO$,
$\therefore △BAE∽ △BDO$,
$\therefore \frac {AE}{OD}=\frac {AB}{BD}$,即$\frac {AE}{3}=\frac {2}{4}$,
解得$AE=\frac {3}{2}$.