22. (12分)如图,在$\mathrm{Rt}△ ABC$中,$∠ C = 90^{\circ}$,$AD$平分$∠ BAC$交$BC$于点$D$,$O$为$AB$上一点,经过点$A$,$D$的$\odot O$分别交$AB$,$AC$于点$E$,$F$.
(1)求证:$BC$是$\odot O$的切线;
(2)若$BE = 8$,$\sin B = \dfrac{5}{13}$,求$\odot O$的半径;
(3)求证:$AD^{2} = AB · AF$.

22. (1) 证明: 如图①, 连接OD,
则 $ OA = OD $,
$ \therefore ∠ ODA = ∠ OAD $.
$ \because AD $ 是 $ ∠ BAC $ 的平分线,
$ \therefore ∠ OAD = ∠ CAD $.
$ \therefore ∠ ODA = ∠ CAD $.
$ \therefore OD // AC $.
$ \therefore ∠ ODB = ∠ C = 90^{\circ} $.
$ \because OD $ 为半径,
$ \therefore BC $ 是 $ \odot O $ 的切线.
(2) 解: $ \because ∠ BDO = 90^{\circ} $,
$ \therefore \sin B = \frac{OD}{BO} = \frac{OD}{BE + OD} = \frac{5}{13} $.
$ \therefore OD = 5 $.
$ \therefore \odot O $ 的半径为5.
(3) 证明: 连接EF.
$ \because AE $ 是直径,
$ \therefore ∠ AFE = 90^{\circ} = ∠ ACB $.
$ \therefore EF // BC $.
$ \therefore ∠ AEF = ∠ B $.
又 $ \because ∠ AEF = ∠ ADF $,
$ \therefore ∠ B = ∠ ADF $.
又 $ \because ∠ OAD = ∠ CAD $,
$ \therefore △ DAB ∽ △ FAD $.
$ \therefore \frac{AD}{AB} = \frac{AF}{AD} $.
$ \therefore AD^{2} = AB · AF $.