16. (2025·南通期中)如图，在$□ ABCD$中，$O$为边$AD$的中点，连接$BO$并延长，交$CD$的延长线于点$E$，连接$AE$，$BD$，$\angle BDC = 90^{\circ}$。
(1) 求证：四边形$ABDE$是矩形；
(2) 连接$OC$，若$AB = 4$，$BD = 4\sqrt{5}$，求$OC$的长。
(1) 求证：四边形$ABDE$是矩形；
(2) 连接$OC$，若$AB = 4$，$BD = 4\sqrt{5}$，求$OC$的长。
答案：
16.(1)$\because O$为$AD$的中点，$\therefore AO = DO.\because$四边形$ABCD$是平行四边形，$\therefore AB // CD.\therefore \angle BAO = \angle EDO$.又$\because \angle AOB = \angle DOE$，$\therefore \triangle AOB \cong \triangle DOE.\therefore AB = DE$.又$\because AB // DE$，
$\therefore$四边形$ABDE$是平行四边形.$\because \angle BDC = 90^{\circ},\therefore \angle BDE = 90^{\circ}.\therefore$四边形$ABDE$是矩形 (2) 如图，过点$O$作$OF ⊥ DE$于点$F.\because$四边形$ABDE$是矩形，$\therefore DE = AB = 4$，$OD = \frac{1}{2}AD$，
$OB = OE = \frac{1}{2}BE$，$AD = BE.\therefore OD = OE.\because OF ⊥ DE$，
$\therefore DF = EF = \frac{1}{2}DE = 2.\therefore OF$为$\triangle BDE$的中位线.$\therefore OF = \frac{1}{2}BD = 2\sqrt{5}.\because$四边形$ABCD$是平行四边形，$\therefore CD = AB = 4$.
$\therefore CF = CD + DF = 6$.在$Rt\triangle OCF$中，由勾股定理，得$OC = \sqrt{CF^{2} + OF^{2}} = \sqrt{6^{2} + (2\sqrt{5})^{2}} = 2\sqrt{14}$
16.(1)$\because O$为$AD$的中点，$\therefore AO = DO.\because$四边形$ABCD$是平行四边形，$\therefore AB // CD.\therefore \angle BAO = \angle EDO$.又$\because \angle AOB = \angle DOE$，$\therefore \triangle AOB \cong \triangle DOE.\therefore AB = DE$.又$\because AB // DE$，
$\therefore$四边形$ABDE$是平行四边形.$\because \angle BDC = 90^{\circ},\therefore \angle BDE = 90^{\circ}.\therefore$四边形$ABDE$是矩形 (2) 如图，过点$O$作$OF ⊥ DE$于点$F.\because$四边形$ABDE$是矩形，$\therefore DE = AB = 4$，$OD = \frac{1}{2}AD$，
$OB = OE = \frac{1}{2}BE$，$AD = BE.\therefore OD = OE.\because OF ⊥ DE$，
$\therefore DF = EF = \frac{1}{2}DE = 2.\therefore OF$为$\triangle BDE$的中位线.$\therefore OF = \frac{1}{2}BD = 2\sqrt{5}.\because$四边形$ABCD$是平行四边形，$\therefore CD = AB = 4$.
$\therefore CF = CD + DF = 6$.在$Rt\triangle OCF$中，由勾股定理，得$OC = \sqrt{CF^{2} + OF^{2}} = \sqrt{6^{2} + (2\sqrt{5})^{2}} = 2\sqrt{14}$
17. (2025·如皋期末)四边形$ABCD$为正方形，$E$为对角线$AC$上一点，连接$DE$。过点$E$作$EF ⊥ DE$，交射线$BC$于点$F$。
(1) 如图①，若点$F$在边$BC$上，求证：$DE = EF$。
(2) 以$DE$，$EF$为邻边作矩形$DEFG$，连接$CG$。
① 如图②，若$AB = 4$，$CE = 3\sqrt{2}$，求$CG$的长；
② 当线段$DE$与正方形$ABCD$一边的夹角是$35^{\circ}$时，请求出$\angle EFC$的度数。
(1) 如图①，若点$F$在边$BC$上，求证：$DE = EF$。
(2) 以$DE$，$EF$为邻边作矩形$DEFG$，连接$CG$。
① 如图②，若$AB = 4$，$CE = 3\sqrt{2}$，求$CG$的长；
② 当线段$DE$与正方形$ABCD$一边的夹角是$35^{\circ}$时，请求出$\angle EFC$的度数。
答案：
17.(1) 连接$BE.\because$四边形$ABCD$是正方形，$\therefore BC = DC$，$\angle BCA = \angle DCA = 45^{\circ}$.又$\because EC = EC$，$\therefore \triangle BCE \cong \triangle DCE$.
$\therefore BE = DE$，$\angle EBC = \angle EDC.\because$四边形$ABCD$是正方形，
$\therefore \angle DCF = 90^{\circ}.\because DE ⊥ EF$，$\therefore \angle DEF = 90^{\circ}.\therefore \angle CDE + \angle CFE = 360^{\circ} - (\angle DCF + \angle DEF) = 180^{\circ}.\because \angle CFE + \angle EFB = 180^{\circ}$，$\therefore \angle CDE = \angle EFB.\therefore \angle EBF = \angle EFB$.
$\therefore BE = EF.\therefore DE = EF$ (2) ①$\because$四边形$DEFG$为矩形，由(1)知，$DE = EF$，$\therefore$四边形$DEFG$为正方形.$\therefore DE = DG$.
$\because$四边形$ABCD$为正方形，$\therefore AD = DC$，$\angle ADC = 90^{\circ} = \angle EDG.\therefore \angle ADE = \angle CDG.\therefore \triangle ADE \cong \triangle CDG.\therefore AE = CG.\because AB = 4$，$\therefore$易得$AC = 4\sqrt{2}.\because CE = 3\sqrt{2}$，$\therefore CG = AE = AC - CE = \sqrt{2}$ ② 当$\angle ADE = 35^{\circ}$时，$\angle CDE = 90^{\circ} - \angle ADE = 55^{\circ}.\because$易知$\angle CDE + \angle EFC = 180^{\circ}$，$\therefore \angle EFC = 125^{\circ}$.当$\angle CDE = 35^{\circ}$时，设$EF$交$DC$于点$H$.
$\because \angle DEH = \angle HCF = 90^{\circ}$，$\angle DHE = \angle CHF$，$\therefore \angle EFC = \angle CDE = 35^{\circ}$.综上所述，$\angle EFC = 125^{\circ}$或$35^{\circ}$
17.(1) 连接$BE.\because$四边形$ABCD$是正方形，$\therefore BC = DC$，$\angle BCA = \angle DCA = 45^{\circ}$.又$\because EC = EC$，$\therefore \triangle BCE \cong \triangle DCE$.
$\therefore BE = DE$，$\angle EBC = \angle EDC.\because$四边形$ABCD$是正方形，
$\therefore \angle DCF = 90^{\circ}.\because DE ⊥ EF$，$\therefore \angle DEF = 90^{\circ}.\therefore \angle CDE + \angle CFE = 360^{\circ} - (\angle DCF + \angle DEF) = 180^{\circ}.\because \angle CFE + \angle EFB = 180^{\circ}$，$\therefore \angle CDE = \angle EFB.\therefore \angle EBF = \angle EFB$.
$\therefore BE = EF.\therefore DE = EF$ (2) ①$\because$四边形$DEFG$为矩形，由(1)知，$DE = EF$，$\therefore$四边形$DEFG$为正方形.$\therefore DE = DG$.
$\because$四边形$ABCD$为正方形，$\therefore AD = DC$，$\angle ADC = 90^{\circ} = \angle EDG.\therefore \angle ADE = \angle CDG.\therefore \triangle ADE \cong \triangle CDG.\therefore AE = CG.\because AB = 4$，$\therefore$易得$AC = 4\sqrt{2}.\because CE = 3\sqrt{2}$，$\therefore CG = AE = AC - CE = \sqrt{2}$ ② 当$\angle ADE = 35^{\circ}$时，$\angle CDE = 90^{\circ} - \angle ADE = 55^{\circ}.\because$易知$\angle CDE + \angle EFC = 180^{\circ}$，$\therefore \angle EFC = 125^{\circ}$.当$\angle CDE = 35^{\circ}$时，设$EF$交$DC$于点$H$.
$\because \angle DEH = \angle HCF = 90^{\circ}$，$\angle DHE = \angle CHF$，$\therefore \angle EFC = \angle CDE = 35^{\circ}$.综上所述，$\angle EFC = 125^{\circ}$或$35^{\circ}$