12. [2025 南通] 如图，网格图中每个小正方形的面积都为1.经过网格点A的一条直线，把网格图分成了两个部分，其中$△ BMN$的面积为3，则$\sin∠ MNB$的值为.

（第12题） （第13题）

12.$\dfrac{\sqrt{6}-\sqrt{2}}{4}$ 【点拨】如图,在图中标注$C,D$,设$NC=x. \because AD// NB,\therefore ∠ MAD=∠ ANC$. 又$\because ∠ MDA=∠ ACN,\therefore △ ANC∽△ MAD. \therefore \dfrac{NC}{AD}=\dfrac{AC}{MD}$. 易知$AC=AD=1,\therefore MD=\dfrac{1}{x}$.$\because △ BMN$的面积为3,网格图中每个小正方形的面积都是1,$\therefore S_{△ AMD}+S_{△ ANC}=3-1=2. \therefore \dfrac{1}{2}MD· AD+\dfrac{1}{2}NC· AC=2. \therefore \dfrac{1}{2}× \dfrac{1}{x}× 1+\dfrac{1}{2}x× 1=2,\therefore x+\dfrac{1}{x}=4$,解得$x_1=2+\sqrt{3},x_2=2-\sqrt{3}$(舍去). $\because AN^2=AC^2+NC^2,\therefore AN^2=1+(2+\sqrt{3})^2=8+4\sqrt{3}=(\sqrt{2}+\sqrt{6})^2,\therefore AN=\sqrt{2}+\sqrt{6}$.$\therefore \sin∠ MNB=\sin∠ ANC=\dfrac{1}{\sqrt{2}+\sqrt{6}}=\dfrac{\sqrt{6}-\sqrt{2}}{4}$,故答案为$\dfrac{\sqrt{6}-\sqrt{2}}{4}$.

13. 如图，已知点$A(5,4)$，点$B$在$y$轴上，点$C(x,0)$且$0＜x＜5$，$BC⊥ AC$于点$C$，连接$AB$，若$AB$与$y$轴正半轴所夹的角为$α$，当$\sinα$取最大值时，对应的$x$值为。

13.$\dfrac{5}{2}$ 【点拨】如图,过点$A$作$AD⊥ x$轴于点$D$,作$AE⊥ y$轴于点$E$,$\therefore ∠ ADC=∠ AEB=90°=∠ BOC$. $\therefore$四边形$ADOE$是矩形. $\therefore AD=OE,AE=OD. \because A(5,4),\therefore AD=OE=4,AE=OD=5. \because C(x,0),\therefore OC=x$. 又$\because 0＜ x＜5$,$\therefore DC=OD-OC=5-x$. 设$OB=m$,则$BE=OE-OB=4-m. \because BC⊥ AC,\therefore ∠ BCA=90°. \therefore ∠ DCA=90°-∠ OCB=∠ OBC. \therefore △ OBC∽△ DCA. \therefore \dfrac{OC}{DA}=\dfrac{OB}{DC}. \therefore \dfrac{x}{4}=\dfrac{m}{5-x},\therefore m=-\dfrac{1}{4}x^2+\dfrac{5}{4}x=-\dfrac{1}{4}(x^2-5x)=-\dfrac{1}{4}(x-\dfrac{5}{2})^2+\dfrac{25}{16}$,当$x=\dfrac{5}{2}$时,$m$取得最大值,此时$BE=4-m$取得最小值,根据勾股定理,得$AB^2=AE^2+BE^2=25+BE^2$,故当$BE$取得最小值时,$AB$也取得最小值.$\because \sinα=\dfrac{AE}{AB}=\dfrac{5}{AB},\therefore AB$取得最小值时,$\sinα=\dfrac{5}{AB}$取最大值,故$x=\dfrac{5}{2}$.

14. 如图,已知抛物线$y=ax^2+3x+c$经过$A(0,4)$和$B(\frac{13}{3},-\frac{16}{9})$两点,与$x$轴交于$M,N$两点($N$在$M$的右侧),直线$AB$与$x$轴相交于点$C$,$P$是直线$AB$上方的抛物线上的一个动点,$PD⊥ x$轴交$AB$于点$D$.
(1)该抛物线的表达式为;
(2)若点$P$与点$N$重合,连接$PA$,求$∠ DAP$的正弦值.

14.【解】(1)$y=-x^2+3x+4$ 【点拨】$\because$抛物线$y=ax^2+3x+c$经过$A(0,4)$和$B(\dfrac{13}{3},-\dfrac{16}{9})$两点,
$\therefore\begin{cases}4=c,\\-\dfrac{16}{9}=(\dfrac{13}{3})^2a+3×\dfrac{13}{3}+c,\end{cases}$解得$\begin{cases}a=-1,\\c=4.\end{cases}$
$\therefore$该抛物线的表达式为$y=-x^2+3x+4$.
(2)如图,过点$C$作$CH⊥ AP$于点$H$,
在抛物线$y=-x^2+3x+4$中,令$y=0$,
则$-x^2+3x+4=0$,解得$x_1=-1,x_2=4$,
$\therefore M(-1,0),N(4,0). \therefore ON=OP=4$.
又$\because A(0,4),\therefore OA=4. \therefore AP=\sqrt{OA^2+ON^2}=4\sqrt{2}$.
设直线$AB$的函数表达式为$y=kx+b$.
$\because$直线过$A(0,4)$和$B(\dfrac{13}{3},-\dfrac{16}{9})$两点,
$\therefore\begin{cases}4=b,\\-\dfrac{16}{9}=\dfrac{13}{3}k+b,\end{cases}$解得$\begin{cases}k=-\dfrac{4}{3},\\b=4.\end{cases}$
$\therefore$直线$AB$的函数表达式为$y=-\dfrac{4}{3}x+4$. 令$y=0$,则$0=-\dfrac{4}{3}x+4$,解得$x=3$,$\therefore C(3,0). \therefore OC=3. \therefore PC=OP-OC=4-3=1$,$AC=\sqrt{4^2+3^2}=5. \because S_{△ ACP}=\dfrac{1}{2}CP· OA=\dfrac{1}{2}PA· CH,\therefore \dfrac{1}{2}× 1× 4=\dfrac{1}{2}× 4\sqrt{2}· CH. \therefore CH=\dfrac{\sqrt{2}}{2}$.
$\therefore \sin∠ DAP=\dfrac{CH}{AC}=\dfrac{\dfrac{\sqrt{2}}{2}}{5}=\dfrac{\sqrt{2}}{10}$.

15. 新考法 构造相似三角形法 如图,在正方形ABCD中,E,F分别是边AD,AB上的点,连接CE,EF,CF.
(1)若正方形ABCD的边长为2,E是AD的中点.
①如图①,当∠FEC=90°时,求证:△AEF∽△DCE;
②如图②,当$\tan∠FCE=\frac{2}{3}$时,AF=;
(2)如图③,延长CF,DA交于点G,当GE=DE, $\sin∠FCE=\frac{1}{3}$时,求证:AE=AF.

15.(1)①【证明】$\because$四边形$ABCD$是正方形,
$\therefore ∠ A=∠ D=90°. \therefore ∠ ECD+∠ CED=90°$.
$\because ∠ CEF=90°,\therefore ∠ AEF+∠ CED=90°$.
$\therefore ∠ AEF=∠ ECD. \therefore △ AEF∽△ DCE$.
②$\dfrac{6}{7}$ 【点拨】如图①,延长$DA$交$CF$的延长线于点$G$,过点$G$作$GH⊥ CE$交$CE$的延长线于点$H. \because ∠ H=∠ D=90°,∠ GEH=∠ CED,\therefore △ GEH∽△ CED. \therefore \dfrac{GH}{CD}=\dfrac{EH}{DE}$.
$\because E$是$AD$的中点,$AD=2$,$\therefore AE=ED=1$. 又$\because CD=2$,$\therefore CD=2DE$,$CE=\sqrt{DE^2+CD^2}=\sqrt{1^2+2^2}=\sqrt{5}. \therefore GH=2HE$. 设$EH=m$,则$GH=2m$,$CH=m+\sqrt{5}. \because \tan∠ FCE=\dfrac{GH}{CH}=\dfrac{2}{3}$,$\therefore \dfrac{2m}{m+\sqrt{5}}=\dfrac{2}{3}. \therefore m=\dfrac{\sqrt{5}}{2}. \therefore EH=\dfrac{\sqrt{5}}{2}$,$GH=\sqrt{5}$.
$\therefore EG=\sqrt{GH^2+EH^2}=\sqrt{(\sqrt{5})^2+(\dfrac{\sqrt{5}}{2})^2}=\dfrac{5}{2}. \therefore AG=EG-AE=\dfrac{5}{2}-1=\dfrac{3}{2}$,$DG=EG+DE=\dfrac{5}{2}+1=\dfrac{7}{2}. \because AF// CD$,$\therefore △ AFG∽△ DCG. \therefore \dfrac{AF}{CD}=\dfrac{AG}{DG}$,即$\dfrac{AF}{2}=\dfrac{\dfrac{3}{2}}{\dfrac{7}{2}}. \therefore AF=\dfrac{6}{7}$.


(2)【证明】如图②,过点$G$作$GH⊥ CE$交$CE$的延长线于点$H$.
由题意设$AD=CD=a$,$GE=DE=t$,$EH=x$,$GH=y$,$CE=n$.
$\therefore CH=x+n$,$AG=2t-a$,$AE=a-t$.
$\because ∠ H=∠ D=90°$,$∠ GEH=∠ CED$,
$\therefore △ GEH∽△ CED. \therefore \dfrac{GH}{CD}=\dfrac{EH}{ED}=\dfrac{EG}{EC}$,
即$\dfrac{y}{a}=\dfrac{x}{t}=\dfrac{t}{n}. \therefore x=\dfrac{t^2}{n}$,$y=\dfrac{at}{n}$.
$\because \sin∠ FCE=\dfrac{1}{3}=\dfrac{GH}{CG}$,$\therefore CG=3GH$.
$\therefore$易得$CH=2\sqrt{2}GH. \therefore 2\sqrt{2}y=x+n$.
$\therefore 2\sqrt{2}× \dfrac{at}{n}=\dfrac{t^2}{n}+n. \therefore 2\sqrt{2}at=t^2+n^2$.
在$\mathrm{Rt}△ CDE$中,$n^2=t^2+a^2$,$\therefore 2\sqrt{2}at=2t^2+a^2. \therefore a=\sqrt{2}t$.
$\because AF// CD$,$\therefore △ AFG∽△ DCG$.
$\therefore \dfrac{AF}{CD}=\dfrac{AG}{DG}$,即$\dfrac{AF}{a}=\dfrac{2t-a}{2t}$.
$\therefore AF=\dfrac{a(2t-a)}{2t}=a-\dfrac{a^2}{2t}=a-t. \because AE=a-t$,$\therefore AE=AF$.