8. 如图,在正五边形ABCDE中,M是AB的中点,连接AC,DM交于点N,则∠CND的度数是.

8.$54°$

9. 如图, 六边形 $ ABCDEF $ 是 $ \odot O $ 的内接正六边形, 设正六边形 $ ABCDEF $ 的面积为 $ S_1 $, $ △ ACE $ 的面积为 $ S_2 $, 则 $ \dfrac{S_1}{S_2} = $ .

9.2

10.(2025·徐州一模)如图，$\odot O$的半径为$\sqrt{2}$，正方形$ABCD$内接于$\odot O$，点$E$在$\overset{\frown}{ADC}$上运动，连接$BE$，作$AF\bot BE$，垂足为$F$，连接$CF$，则$CF$长的最小值为________.

10.$\sqrt{5}-1$

11. 如图，$△ ABC$内接于$\odot O$，$AB=AC$，$∠ BAC=36°$，$AB$，$AC$的垂直平分线分别交$\odot O$于点$E$，$F$. 求证:五边形$AEBCF$是$\odot O$的内接正五边形.

11.证明:如答图,连接BF,CE.

$\because AB=AC,∠ BAC=36°$,
$\therefore ∠ ABC=∠ ACB=72°,\overset{\frown}{AB}=\overset{\frown}{AC}$.
$\because AB,AC$的垂直平分线分别交$\odot O$于点E,F,
$\therefore AF=CF,AE=BE,\therefore \overset{\frown}{AF}=\overset{\frown}{CF},\overset{\frown}{AE}=\overset{\frown}{BE}$.
$\because \overset{\frown}{AB}=\overset{\frown}{AC},\therefore \overset{\frown}{AF}=\overset{\frown}{CF}=\overset{\frown}{AE}=\overset{\frown}{BE}$,
易得$∠ BAC=∠ BCE=∠ ACE=∠ ABF=∠ FBC=36°$,
$\therefore BC=BE=AE=AF=FC$,
$\therefore$五边形AEBCF为正五边形.
又$\because$点A,E,B,C,F都在$\odot O$上,
$\therefore$五边形AEBCF是$\odot O$的内接正五边形.

12. 如图,中心为O的正六边形ABCDEF的半径为6 cm,点P,Q同时分别从A,D两点出发,以1 cm/s的速度沿AF,DC向终点F,C运动,连接PB,PE,QB,QE,设运动时间为t(s).
(1)求证:四边形PBQE为平行四边形;
(2)当四边形PBQE是矩形时,求矩形PBQE的面积与正六边形ABCDEF的面积之比.

12.(1)证明:$\because$六边形ABCDEF是正六边形,
$\therefore AB = BC = CD = DE = EF = FA,∠ A = ∠ ABC = ∠ C=∠ D=∠ DEF=∠ F$.
$\because$点P,Q同时分别从A,D两点出发,以1 cm/s的速度沿AF,DC向终点F,C运动,$\therefore AP=DQ=t$ cm.
在$△ ABP$和$△ DEQ$中,$\begin{cases} AB=DE, \\ ∠ A=∠ D, \\ AP=DQ, \end{cases}$
$\therefore △ ABP≌ △ DEQ(\mathrm{SAS}),\therefore BP=EQ$.
同理可证$PE=QB,\therefore$四边形PBQE为平行四边形.
(2)解:如答图①,连接BE,OA,BD,AE,则$∠ AOB=\frac{360°}{6}=60°$.$\because OA=OB,\therefore △ AOB$是等边三角形,
$\therefore AB=OA=6$ cm,$BE=2OB=12$ cm.
当$t=0$时,点P与点A重合,点Q与点D重合,四边形PBQE即为四边形ABDE.$\because ∠ EAF=∠ AEF=30°$,
$\therefore ∠ BAE=∠ BAF-∠ FAE=120°-30°=90°$,
$\therefore$此时四边形ABDE是矩形,即四边形PBQE是矩形.
当$t=6$时,点P与点F重合,点Q与点C重合,四边形PBQE即为四边形FBCE,如答图②,同法可知$∠ BFE=90°$,此时四边形PBQE是矩形.
$\because AE=\sqrt{12^2-6^2}=6\sqrt{3}\ (\mathrm{cm}),\therefore S_{\mathrm{矩形}PBQE}=S_{\mathrm{矩形}ABDE}=AB· AE=6×6\sqrt{3}=36\sqrt{3}\ (\mathrm{cm}^2)$.
$\because S_{\mathrm{正六边形}ABCDEF}=6S_{△ AOB}=6×\frac{1}{4}S_{\mathrm{矩形}ABDE}=6×\frac{1}{4}×36\sqrt{3}=54\sqrt{3}\ (\mathrm{cm}^2),\therefore$矩形PBQE的面积与正六边形ABCDEF的面积之比为$\frac{2}{3}$.