5. 如图,以 AB 为直径的$\odot O$经过$△ ABC$的顶点 C,点 E 是$△ ABC$的内心,连接 AE 并延长交$\odot O$于点 D,连接 BE,BD.
(1) 试判断$△ BDE$的形状并证明;
(2) 若$BE=5\sqrt{2},∠BAC=60°$,求图中阴影部分的面积.

(1) 解:$△ BDE$是等腰直角三角形,证明:$\because AB$为直径,$\therefore ∠ ADB=90°$. $\because$ 点 E 是$△ ABC$的内心,$\therefore AE,BE$分别平分$∠ BAC、∠ ABC$,$\therefore ∠ BAE=∠ CAD$,$∠ ABE=∠ CBE$. $\because ∠ CAD=∠ CBD$,$\therefore ∠ CBD=∠ BAE$. $\because ∠ DBE=∠ CBD+∠ CBE$,$∠ BED=∠ ABE+∠ BAE$,$\therefore ∠ DBE=∠ BED$,$\therefore BD=DE$,$\therefore △ BDE$是等腰直角三角形.
(2) 如图,连接$OC,OD$,$OD$与$BC$交于点$F$,由(1)可知$△ BDE$是等腰直角三角形,$\therefore BD^2+DE^2=BE^2$,$\therefore BD^2+BD^2=(5\sqrt{2})^2$,$\therefore BD=5$. $\because AE$平分$∠ BAC$,$∠ BAC=60°$,$\therefore ∠ BAD=∠ CAD=\frac{1}{2}∠ BAC=\frac{1}{2}×60°=30°$,$∠ BOC=2∠ BAC=2×60°=120°$,$\therefore ∠ BOD=∠ COD=2∠ BAD=2×30°=60°$. $\because OB=OD=OC$,$\therefore △ OBD$为等边三角形,$OD⊥ BC$,$\therefore OB=BD=5$. $\because OB=OC$,$\therefore ∠ OBC=∠ OCB=30°$,$\therefore OF=\frac{1}{2}OB=\frac{5}{2}$,$\therefore BF=\sqrt{OB^2-OF^2}=\sqrt{5^2-(\frac{5}{2})^2}=\frac{5}{2}\sqrt{3}$,$\therefore BC=2BF=5\sqrt{3}$,$\therefore S_{\mathrm{阴影部分}}=S_{\mathrm{扇形}OBC}-S_{△ OBC}=\frac{120π×5^2}{360}-\frac{1}{2}· BC· OF=\frac{25π}{3}-\frac{1}{2}×5\sqrt{3}×\frac{5}{2}=\frac{25π}{3}-\frac{25}{4}\sqrt{3}$.

6. 如图△ABC内接于⊙O，∠B＝60°，CD是⊙O的直径，点P是CD延长线上一点，且AP＝AC，PD＝3.
（1）求证：PA是⊙O的切线；
（2）求⊙O的直径；
（3）当点B在CD下方运动时，直接写出△ABC内心的运动路线长是.

(1) 证明:如图①,连接$AO,AD$,$\because CD$是圆 O 的直径,$\therefore ∠ CAD=90°$. $\because ∠ B=60°$,$\therefore ∠ ADC=60°$. $\because AO=DO$,$\therefore △ AOD$是等边三角形,$\therefore ∠ OAD=60°$. $\because AP=AC$,$\because CD$是$\odot O$直径,$\therefore ∠ CAD=90°$,$\therefore ∠ P=∠ ACP=30°$,$\therefore ∠ PAD=30°$,$\therefore ∠ PAO=30°+60°=90°$,$\therefore AO⊥ PA$. $\because A$点在圆上,$\therefore PA$是$\odot O$的切线.
(2) 由(1)可知,$∠ P=∠ PAD=30°$,$\therefore PD=AD$. $\because PD=3$,$\therefore AD=3$. $\because ∠ ACD=30°$,$∠ CAD=90°$,$\therefore CD=2AD=6$,$\therefore \odot O$的直径是6.
(3) 设$△ ABC$的内切圆圆心为 M,如图②,连接$AM,CM,BM$,$\because ∠ ABC=60°$,$\therefore ∠ BAC+∠ BCA=120°$. $\because AM$是$∠ BAC$的平分线,$CM$是$∠ BCA$的平分线,$\therefore ∠ MAC+∠ MCA=60°$,$\therefore ∠ AMC=120°$. 由(2)可知,$AC=\sqrt{CD^2-AD^2}=3\sqrt{3}$,$\therefore$ 点 M 在以 AC 为弦,AC 弦所对的圆周角为$120°$的圆上. 作$△ AMC$的外接圆,圆心为点 N,连接$AN,CN$,$\because ∠ AMC=120°$,$\therefore ∠ ANC=120°$,$\therefore ∠ ABC+∠ ANC=180°$,$\therefore$ 点 N 在$\odot O$上. 连接$ON$,$\because AN=CN$,$\therefore ∠ CON=∠ ABC=60°$,$\therefore △ OCN$是等边三角形,$\therefore CN=OC=3$. 当点 B 与点 D 重合时,$∠ CNB=90°$. $\therefore △ ABC$内心的运动路线长是$\frac{1}{4}×2π×3=\frac{3}{2}π$.