20. (2025·南通期末)学校航模小组打算制作模型飞机，设计了如图所示的模型飞机的机翼图纸. 已知$ AB// CD$，且$ AB$，$ CD$均与水平方向垂直，机翼前缘$ AC$、机翼后缘$ BD$与水平方向形成的夹角度数分别为$45°$，$30°$. 若$ AB=20\ cm$，点$ D$到直线$ AB$的距离$ DG=30\ cm$，求$ CD$的长(结果保留根号).
答案：
20.如图，过点A作$AE\perp DC$，交DC的延长线于点E，过点B作$BF\perp ED$于点F.由题意，可得$\angle E=\angle ABF=\angle BFE=\angle BFD=\angle GDF=\angle G=90^{\circ}$，$\angle EAC=45^{\circ}$，$\angle FBD=30^{\circ}$，则四边形ABFE、四边形BGDF均为矩形，$\therefore EF=AB=20$cm，$AE=BF=GD=30$cm.在$Rt\triangle BDF$中，$\tan\angle FBD=\frac{FD}{BF}$，则$FD=BF·\tan\angle FBD=30×\tan30^{\circ}=30×\frac{\sqrt{3}}{3}=10\sqrt{3}(cm)$，$\therefore ED=EF+FD=(20+10\sqrt{3})cm$.在$Rt\triangle AEC$中，$\angle EAC=45^{\circ}$，则易得$EC=AE=30$cm，$\therefore CD=ED-EC=20+10\sqrt{3}-30=(10\sqrt{3}-10)cm.\therefore CD$的长为$(10\sqrt{3}-10)cm$
20.如图，过点A作$AE\perp DC$，交DC的延长线于点E，过点B作$BF\perp ED$于点F.由题意，可得$\angle E=\angle ABF=\angle BFE=\angle BFD=\angle GDF=\angle G=90^{\circ}$，$\angle EAC=45^{\circ}$，$\angle FBD=30^{\circ}$，则四边形ABFE、四边形BGDF均为矩形，$\therefore EF=AB=20$cm，$AE=BF=GD=30$cm.在$Rt\triangle BDF$中，$\tan\angle FBD=\frac{FD}{BF}$，则$FD=BF·\tan\angle FBD=30×\tan30^{\circ}=30×\frac{\sqrt{3}}{3}=10\sqrt{3}(cm)$，$\therefore ED=EF+FD=(20+10\sqrt{3})cm$.在$Rt\triangle AEC$中，$\angle EAC=45^{\circ}$，则易得$EC=AE=30$cm，$\therefore CD=ED-EC=20+10\sqrt{3}-30=(10\sqrt{3}-10)cm.\therefore CD$的长为$(10\sqrt{3}-10)cm$
21. 在正方形$ ABCD$中，点$ P$在边$ BC$上(不与端点$ B$，$ C$重合)，点$ B$关于直线$ AP$的对称点为$ E$，$ BE$与$ AP$交于点$ O$.
(1)如图①，连接$ DE$，则$\angle BED$的度数为
(2)如图②，若$ AB=4$，$ P$是$ BC$的中点，连接$ CE$，求$ CE$的长.
(3)如图③，过点$ D$作$ DF\perp BE$，交直线$ BE$于点$ F$，连接$ OC$，$ CF$. 若$ OC\perp CF$，求$\tan \angle BAP$的值.
(1)如图①，连接$ DE$，则$\angle BED$的度数为
$135^{\circ}$
.(2)如图②，若$ AB=4$，$ P$是$ BC$的中点，连接$ CE$，求$ CE$的长.
(3)如图③，过点$ D$作$ DF\perp BE$，交直线$ BE$于点$ F$，连接$ OC$，$ CF$. 若$ OC\perp CF$，求$\tan \angle BAP$的值.
答案：
21.(1)$135^{\circ}$ 解析：如图①，连接AE.$\because$点B关于直线AP的对称点为E，$\therefore AB=AE.\because$四边形ABCD是正方形，$\therefore AB=AD$，$\angle BAD=90^{\circ}.\therefore AB=AE=AD.\therefore\angle ABE=\angle AEB$，$\angle AED=\angle ADE.\therefore\angle ABE+\angle ADE=\angle AEB+\angle AED.\because\angle BAD+\angle ABE+\angle AEB+\angle AED+\angle ADE=360^{\circ}$，即$90^{\circ}+2(\angle AEB+\angle AED)=360^{\circ}$，$\therefore\angle AEB+\angle AED=135^{\circ}$，即$\angle BED$的度数为$135^{\circ}$.
(2)如图②，连接PE.$\because$四边形ABCD是正方形，$\therefore AB=BC=4.\because P$是BC的中点，$\therefore BP=PC=\frac{1}{2}BC=2.\because$点B关于直线AP的对称点为E，$\therefore BP=PE$，$BE\perp AP$，$BO=OE.\therefore BP=PE=PC.\therefore$易得$\angle BEC=90^{\circ}.\because AB=4$，$BP=2$，$\therefore AP=\sqrt{AB^{2}+BP^{2}}=\sqrt{4^{2}+2^{2}}=2\sqrt{5}.\because\cos\angle APB=\frac{BP}{AP}=\frac{OP}{BP}$，$\therefore\frac{2}{2\sqrt{5}}=\frac{OP}{2}.\therefore OP=\frac{2\sqrt{5}}{5}.\because BO=OE$，$BP=PC$，$\therefore CE=2OP=\frac{4\sqrt{5}}{5}$
(3)如图③，设BF与CD交于点H.
$\because DF\perp BF$，$OC\perp CF$，四边形ABCD是正方形，$\therefore\angle DFB=90^{\circ}=\angle BCD=\angle OCF.\therefore\angle BCD-\angle OCD=\angle OCF-\angle OCD$，即$\angle BCO=\angle DCF$.又$\because\angle BHC=\angle DHF$，$\angle CBO=\angle CDF$.又$\because BC=DC$，$\therefore\triangle BCO\cong\triangle DCF.\therefore BO=DF$.又$\because\angle BOP=\angle DFH=90^{\circ}$，$\angle CBO=\angle CDF$，$\therefore\triangle BOP\cong\triangle DFH.\therefore BP=DH.\because\angle CBH+\angle BPA=90^{\circ}=\angle BPA+\angle BAP$，$\therefore\angle BAP=\angle CBH$.又$\because AB=BC$，$\angle ABP=\angle BCH=90^{\circ}$，$\therefore\triangle ABP\cong\triangle BCH.\therefore BP=CH$.
$\therefore BP=CH=DH.\therefore BP=\frac{1}{2}CD.\because$四边形ABCD是正方形，$\therefore\angle ABP=90^{\circ}$，$AB=CD.\therefore BP=\frac{1}{2}AB.\therefore$在$Rt\triangle ABP$中，$\tan\angle BAP=\frac{BP}{AB}=\frac{1}{2}$
21.(1)$135^{\circ}$ 解析：如图①，连接AE.$\because$点B关于直线AP的对称点为E，$\therefore AB=AE.\because$四边形ABCD是正方形，$\therefore AB=AD$，$\angle BAD=90^{\circ}.\therefore AB=AE=AD.\therefore\angle ABE=\angle AEB$，$\angle AED=\angle ADE.\therefore\angle ABE+\angle ADE=\angle AEB+\angle AED.\because\angle BAD+\angle ABE+\angle AEB+\angle AED+\angle ADE=360^{\circ}$，即$90^{\circ}+2(\angle AEB+\angle AED)=360^{\circ}$，$\therefore\angle AEB+\angle AED=135^{\circ}$，即$\angle BED$的度数为$135^{\circ}$.
(2)如图②，连接PE.$\because$四边形ABCD是正方形，$\therefore AB=BC=4.\because P$是BC的中点，$\therefore BP=PC=\frac{1}{2}BC=2.\because$点B关于直线AP的对称点为E，$\therefore BP=PE$，$BE\perp AP$，$BO=OE.\therefore BP=PE=PC.\therefore$易得$\angle BEC=90^{\circ}.\because AB=4$，$BP=2$，$\therefore AP=\sqrt{AB^{2}+BP^{2}}=\sqrt{4^{2}+2^{2}}=2\sqrt{5}.\because\cos\angle APB=\frac{BP}{AP}=\frac{OP}{BP}$，$\therefore\frac{2}{2\sqrt{5}}=\frac{OP}{2}.\therefore OP=\frac{2\sqrt{5}}{5}.\because BO=OE$，$BP=PC$，$\therefore CE=2OP=\frac{4\sqrt{5}}{5}$
(3)如图③，设BF与CD交于点H.
$\because DF\perp BF$，$OC\perp CF$，四边形ABCD是正方形，$\therefore\angle DFB=90^{\circ}=\angle BCD=\angle OCF.\therefore\angle BCD-\angle OCD=\angle OCF-\angle OCD$，即$\angle BCO=\angle DCF$.又$\because\angle BHC=\angle DHF$，$\angle CBO=\angle CDF$.又$\because BC=DC$，$\therefore\triangle BCO\cong\triangle DCF.\therefore BO=DF$.又$\because\angle BOP=\angle DFH=90^{\circ}$，$\angle CBO=\angle CDF$，$\therefore\triangle BOP\cong\triangle DFH.\therefore BP=DH.\because\angle CBH+\angle BPA=90^{\circ}=\angle BPA+\angle BAP$，$\therefore\angle BAP=\angle CBH$.又$\because AB=BC$，$\angle ABP=\angle BCH=90^{\circ}$，$\therefore\triangle ABP\cong\triangle BCH.\therefore BP=CH$.
$\therefore BP=CH=DH.\therefore BP=\frac{1}{2}CD.\because$四边形ABCD是正方形，$\therefore\angle ABP=90^{\circ}$，$AB=CD.\therefore BP=\frac{1}{2}AB.\therefore$在$Rt\triangle ABP$中，$\tan\angle BAP=\frac{BP}{AB}=\frac{1}{2}$