【教材$P_{124}T_{13}$变式 1】 如图,$AB为\odot O$的直径,$C为\odot O$上一点,$I为\triangle ABC$的内心,$CI的延长线交\odot O于点D$,连接$AD$.
(1)求证:$DA= DI$;
(2)若$AC= 8$,$BC= 6$.求$ID$的长.

解:(1)连接 AI.
∵AB 为⊙O 的直径,
∴∠ACB = 90°,
∵I 为△ABC 的内心,
∴∠BAI = ∠CAI,
∠ACD = ∠BCD = ∠BAD = 45°.
∵∠DAI = ∠BAD + ∠BAI,
∠DIA = ∠ACD + ∠CAI,
∴∠DAI = ∠DIA,∴DA = DI;
(2)连接 BD.∵AB 为⊙O 的直径,
∴∠ACB = ∠ADB = 90°,
AB = $\sqrt{AC^{2}+BC^{2}} = 10$.
∵∠ABD = ∠ACD = ∠BCD = ∠BAD,∴AD = BD,
∴△ABD 是等腰直角三角形,
∴ID = AD = $\frac{\sqrt{2}}{2}AB = 5\sqrt{2}$.

【教材$P_{124}T_{13}$变式 2】 如图,$AB是\odot O$的直径,$D为\odot O$上一点,$E为\triangle ABD$的内心,连接$AE$,$OE$,$\angle B= 60^{\circ}$.
(1)求证:$AE= CE$;
(2)若$BD= 4$,求$OE$的长.

解:(1)连接 AC,
则∠C = ∠B = 60°.
又∵∠ADC = 45°,
∠DAE = ∠BAE = 15°,
∴∠AEC = ∠DAE + ∠ADC = 60°,
同上可证 AC = CE,
∴△AEC 是等边三角形,
∴AE = CE;
(2)连接 OC,BC.∵AB = 2BD = 8,
∴BC = AC = $\frac{\sqrt{2}}{2}AB = 4\sqrt{2}$.
延长 EO,交 AC 于点 F.
∵EA = EC,OA = OC,
∴EF 垂直平分 AC,
∴AF = CF = OF = $2\sqrt{2}$.
又∵EF = $\sqrt{3}AF = 2\sqrt{6}$,
∴OE = EF - OF = $2\sqrt{6} - 2\sqrt{2}$.

【教材$P_{124}T_{13}$变式 3】 如图,$BC为\triangle ABC的外接圆\odot O$的直径,点$M为\triangle ABC$的内心,$AM的延长线交\odot O于点D$.
(1)求证:$BC= \sqrt{2}DM$;
(2)若$DM= 5\sqrt{2}$,$AB= 8$,求$AM和BM$的长.

解:(1)连接 DB,DC,
易证 BC = $\sqrt{2}BD$,
BD = DM,∴BC = $\sqrt{2}DM$;
(2)∵BC = $\sqrt{2}DM = 10$,∴AC = 6.
过点 M 作 ME⊥AB 于点 E,
∴ME = AE = $\frac{AB + AC - BC}{2} = 2$,
∴AM = $\sqrt{2}ME = 2\sqrt{2}$.
∵BE = AB - AE = 8 - 2 = 6,
∴BM = $\sqrt{BE^{2}+EM^{2}}$
= $\sqrt{6^{2}+2^{2}}$
= $2\sqrt{10}$.