15. 如图,在三角形 BCD 中,E 为直线 BC 上一点,将线段 CD 沿 CB 方向平移至 EA,直线 BD 与直线 AE 交于点 O,N 是直线 CD 上一点.若$\frac{BO}{DO}=\frac{2}{3}$,三角形 AOD 与三角形 BOE 面积和为 13,当 BN 的长的最小值为 10 时,则 AE 的长为.

15. 5或$\dfrac{1}{5}$ 【点拨】本题考查三角形的面积,垂线段最短,平移的性质.
【解析】如题图,当点O在BD上时,连接AB,DE.根据平移的性质,得$AD // CE$,$AD = CE$,$AE = CD$,$\therefore S_{△ ADE} = S_{△ ADB}$,$\therefore S_{△ DOE} = S_{△ AOB}$.$\because \dfrac{BO}{DO} = \dfrac{2}{3}$,$\therefore \dfrac{S_{△ AOB}}{S_{△ AOD}} = \dfrac{2}{3}$,$\dfrac{S_{△ BOE}}{S_{△ DOE}} = \dfrac{2}{3}$,$\therefore \dfrac{S_{△ BOE}}{S_{△ AOB}} = \dfrac{2}{3}$.设$S_{△ BOE} = 4m$,则$S_{△ AOB} = S_{△ DOE} = 6m$,则$S_{△ AOD} = 9m$,$\therefore S_{△ BOE} + S_{△ AOD} = 13m = 13$,$\therefore m = 1$,$\therefore S_{△ BOE} = 4$,$S_{△ DOE} = 6$,$S_{△ AOD} = 9$,$\therefore S_{△ BDE} = 10$,$S_{△ ADE} = 15$,易得$S_{△ CDE} = S_{△ ADE} = 15$,$\therefore S_{△ BCD} = 25$.当$BN ⊥ CD$时,BN最小,$\therefore$ 此时,$S_{△ BCD} = \dfrac{1}{2} CD × BN = 5CD = 25$,$\therefore CD = 5$,$\therefore AE = CD = 5$;如图,当点O在DB的延长线上时,连接AB,DE.则同理可推出$S_{△ BOE} = 4$,$S_{△ DOE} = 6$,$S_{△ AOD} = 9$,$S_{△ BDE} = 2$,$S_{△ ADB} = S_{△ ADE} = S_{△ CDE} = 3$,$\therefore S_{△ BCD} = 1$.当$BN ⊥ CD$时,BN最小,$\therefore$ 此时,$S_{△ BCD} = \dfrac{1}{2} CD × BN = 5CD = 1$,$\therefore CD = \dfrac{1}{5}$,$\therefore AE = CD = \dfrac{1}{5}$.综上所述,AE的长为5或$\dfrac{1}{5}$.故答案为5或$\dfrac{1}{5}$.

16. 一副三角尺按如图所示摆放在量角器上,边$PD$与量角器的$0°$刻度线重合,边$AP$与量角器的$180°$刻度线重合.将三角尺$PCD$绕点$P$以每秒$3°$的速度逆时针旋转,同时三角尺$ABP$绕点$P$以每秒$2°$的速度顺时针旋转,当三角尺$PCD$的$PC$边与$180°$刻度线重合时两块三角尺都停止运动,则当运动时间$t=$秒时,两块三角尺有一组边平行.

16. 6或9或15或33 【点拨】本题考查平行线的性质,旋转的性质.
【解析】①当$AP // CD$时,$∠ APD + ∠ D = 180°$.$\because ∠ D = 30°$,$\therefore ∠ APD = 150°$,$\therefore 180° - 5t° = 150°$,解得$t=6$;②当$AB // PD$时,$∠ A + ∠ APD = 180°$.$\because ∠ A = 45°$,$\therefore ∠ APD = 135°$,$\therefore 180° - 5t° = 135°$,解得$t=9$;③当$AB // CD$时,$∠ APD = 105° = 180° - 5t°$,解得$t=15$;④当$AB // CP$时,$∠ CPB = 90°$,$\therefore ∠ APD = 60° + 45° - 90° = 180° - 5t°$,解得$t=33$;⑤当$AP // CD$时,$∠ C + ∠ APC = 180°$,$\therefore ∠ APC = 90°$,$\therefore ∠ APD = 30° = 5t° - 180°$,解得$t=42 ＞ \dfrac{180-60}{3}=40$(舍去).综上所述,当运动时间$t=6$或9或15或33秒时,两块三角尺有一组边平行.故答案为6或9或15或33.

17. (8分)根据条件进行推理,得出结论,并在括号内注明理由.
如图,$EF // AD$,$EF // BC$,$CF$平分$∠ ACE$,$∠ DAC = 125°$,$∠ ACF = 15°$.求$∠ FEC$的度数.
解:$\because EF // AD$,$EF // BC$(),
$\therefore AD //$(),
$\therefore ∠ DAC +$$= 180°$().
$\because ∠ DAC = 125°$,
$\therefore ∠ ACB = 180° - ∠ DAC = 55°$.
又$\because CF$平分$∠ ACE$,$∠ ACF = 15°$(已知),
$\therefore ∠ ACE = 2∠ ACF = 30°$(),
$\therefore ∠ BCE = ∠ ACB - ∠ ACE = 25°$.
$\because EF // BC$,
$\therefore ∠ FEC =$$= 25°$().

17. 【点拨】本题考查平行线的性质和角平分线的定义.
【解析】$\because EF // AD$,$EF // BC$(已知),
$\therefore AD // BC$(平行于同一直线的两直线互相平行),
$\therefore ∠ DAC + ∠ ACB = 180°$(两直线平行,同旁内角互补).
$\because ∠ DAC = 125°$,
$\therefore ∠ ACB = 180° - ∠ DAC = 55°$.
又$\because CF$平分$∠ ACE$,$∠ ACF = 15°$(已知),
$\therefore ∠ ACE = 2∠ ACF = 30°$(角平分线的定义),
$\therefore ∠ BCE = ∠ ACB - ∠ ACE = 25°$.
$\because EF // BC$,
$\therefore ∠ FEC = ∠ BCE = 25°$(两直线平行,内错角相等).
故答案为已知;$BC$;平行于同一直线的两直线互相平行;$∠ ACB$;两直线平行,同旁内角互补;角平分线的定义;$∠ BCE$;两直线平行,内错角相等.

18. (5分)已知$2a - 1$的平方根是$\pm 3$,$3a + b - 9$的立方根是2,$c$是$\sqrt{17}$的整数部分,求$a + 2b + c$的值.

18. 【点拨】本题考查平方根以及立方根和估算无理数的大小.
【解析】$\because 2a - 1$的平方根是$\pm 3$,$\therefore 2a - 1 = 9$,解得$a=5$.
$\because 3a + b - 9$的立方根是2,$\therefore 15 + b - 9 = 8$,解得$b=2$.
$\because c$是$\sqrt{17}$的整数部分,$\therefore c=4$,则$a + 2b + c = 5 + 2 × 2 + 4 = 13$.