7. (2026·雅安中考)如图,在$△ ABC$中,$∠ B=90°$,$AM$是角平分线,$O$是$AC$上一点,经过点$A$、点$M$的$\odot O$分别交$AB$,$AC$于点$E$,点$F$.
(1)判断$BC$与$\odot O$的位置关系,并说明理由;
(2)求证:$CM^2 = CF · CA$;
(3)若$CF=2$,$\sin C=\frac{3}{5}$,求$AE$的长.

7. (1) BC 与$\odot O$ 的位置关系是相切.理由如下:如图,连接 OM,$\because AM$ 是 $∠ BAC$ 的平分线,$\therefore ∠ BAM=∠ OAM$.$\because OA=OM$,$\therefore ∠ OMA=∠ OAM$,$\therefore ∠ OMA=∠ BAM$,$\therefore OM // AB$.$\because ∠ B=90°$,$\therefore ∠ OMC=∠ B=90°$,即 $OM ⊥ BC$.$\because OM$ 为$\odot O$ 的半径,$\therefore BC$ 与$\odot O$ 的位置关系是相切.
(2) 如图,连接 FM,$\because AF$ 是$\odot O$ 的直径,$\therefore ∠ AMF=90°=∠ B$.$\because AM$ 是 $∠ BAC$ 的平分线,$\therefore ∠ BAM=∠ OAM$,$\therefore ∠ AMB=90°-∠ BAM=90°-∠ OAM=∠ AFM$,$\therefore ∠ AMC=180°-∠ AMB=180°-∠ AFM=∠ MFC$,$\therefore △ CFM ∽ △ CMA$,$\therefore \frac{CF}{CM}=\frac{CM}{CA}$,即 $CM^2=CF· CA$.
(3) 如图,连接 EF,由(1)知 $∠ OMC=90°$,在 $\mathrm{Rt}△ OMC$ 中,$\sin C=\frac{OM}{OC}=\frac{3}{5}$,设 $OM=OF=3a$,则 $OC=5a$,$\therefore CF=OC-OF=2a=2$,解得 $a=1$,$\therefore OF=3$,$AF=2OF=6$.$\because AF$ 为$\odot O$ 的直径,$\therefore ∠ AEF=90°$,$\therefore ∠ B=∠ AEF=90°$,$\therefore EF // BC$,$\therefore ∠ AFE=∠ C$,$\therefore \sin∠ AFE=\frac{AE}{AF}=\frac{3}{5}$,$\therefore AE=\frac{3}{5}AF=\frac{18}{5}$.

8. (2026·新疆中考)如图,△ABC内接于⊙O,AD平分∠BAC交⊙O于点D,连接BD,CD,经过点D的直线与AB的延长线相交于点E,且∠BDE=$\frac{1}{2}$∠BAC.
(1)求证:DE是⊙O的切线;
(2)如果点I是△ABC的内心,$\tan∠BAC=\frac{12}{5}$,BC=24,求DI的值.

8. (1) 如图①,连接 DO 并延长,交 $\odot O$ 于点 F,连接 BF,$\because \overset{\frown}{BD}=\overset{\frown}{BD}$,$\therefore ∠ 1=∠ 3$.$\because DF$ 是$\odot O$ 的直径,$\therefore ∠ FBD=90°$,$\therefore ∠ 1+∠ 2=90°$.$\because AD$ 平分 $∠ BAC$,$\therefore ∠ 3=\frac{1}{2}∠ BAC$.$\because ∠ BDE=\frac{1}{2}∠ BAC$,$\therefore ∠ 3=∠ BDE$,$\therefore ∠ 1=∠ BDE$,$\therefore ∠ BDE+∠ 2=∠ EDF=90°$.$\because OD$ 是$\odot O$ 的半径,$\therefore DE$ 是$\odot O$ 的切线.
(2) 如图②,连接 BI,CI,OB,OC,OD,OD 交 BC 于点 M.$\because AD$ 平分 $∠ BAC$,$\therefore ∠ BAD=∠ CAD$,$\therefore \overset{\frown}{BD}=\overset{\frown}{CD}$,$\therefore ∠ 4=∠ 5$.$\because DO$是$\odot O$ 的半径,$\therefore OD ⊥ BC$,$BM=CM=\frac{1}{2}BC=\frac{1}{2}×24=12$.$\because OB=OC$,$\therefore ∠ BOC=2∠ BOM$.$\because ∠ BOC=2∠ BAC$,$\therefore ∠ BOM=∠ BAC$.$\because \tan∠ BAC=\frac{12}{5}$,$\therefore \tan∠ BOM=\frac{BM}{OM}=\frac{12}{OM}=\frac{12}{5}$,$\therefore OM=5$,$\therefore OB=\sqrt{OM^2+BM^2}=13$,$\therefore OD=13$,$\therefore DM=OD-OM=8$,$\therefore BD=\sqrt{BM^2+DM^2}=\sqrt{12^2+8^2}=4\sqrt{13}$.$\because$ 点 I 是 $△ ABC$ 的内心,$\therefore ∠ ABI=∠ CBI$.$\because \overset{\frown}{BD}=\overset{\frown}{BD}$,$\therefore ∠ 5=∠ BAD$,$\therefore ∠ 4=∠ BAD$.$\because ∠ DBI=∠ 4+∠ CBI$,$∠ 6=∠ BAD+∠ ABI$,$\therefore ∠ DBI=∠ 6$,$\therefore DI=DB=4\sqrt{13}$.