《2025年综合应用创新题典中点七年级数学下册冀教版》

11. （14分）如图，点D，E分别为三角形ABC的边AB，AC上的点，点F，G分别在BC，AB上，∠AED = ∠C，∠DEF = ∠B，∠EFG = 90°. 请说明FG⊥AB.

答案： 【解】$\because\angle AED=\angle C,\therefore DE// BC.\therefore\angle DEF=\angle EFC.\because\angle DEF=\angle B,\therefore\angle EFC=\angle B.\therefore DB// EF.\therefore\angle AGF+\angle EFG = 180^{\circ}.\because\angle EFG = 90^{\circ},\therefore\angle AGF = 90^{\circ}.\therefore FG\perp AB$.

12. （16分）如图，点D，E在AB上，点F，G分别在BC，CA上，且DG//BC，∠1 = ∠2.
（1）请说明DC//EF；
（2）若EF⊥AB，∠1 = 55°，求∠ADG的度数.

答案： 【解】
(1)$\because DG// BC,\therefore\angle 1=\angle DCB.\because\angle 1=\angle 2,\therefore\angle 2=\angle DCB.\therefore DC// EF$.
(2)$\because EF\perp AB,DC// EF,\therefore DC\perp AB.\therefore\angle CDA = 90^{\circ}.\therefore\angle ADG=\angle CDA-\angle 1 = 90^{\circ}-55^{\circ}=35^{\circ}$.

13. （20分）已知：点A在射线CE上，∠C = ∠D.
（1）如图①，若AC//BD，请说明：AD//BC；
（2）如图②，若∠BAC = ∠BAD，BD⊥BC，请探究∠DAE与∠C的数量关系；
（3）如图③，在（2）的条件下，过点D作DF//BC交射线CE于点F，当∠DFE = 8∠DAE时，求∠BAD的度数.

答案： 【解】
(1)$\because AC// BD,\therefore\angle DAE=\angle D$. 又$\because\angle C=\angle D,\therefore\angle DAE=\angle C.\therefore AD// BC$.
(2)$\angle DAE + 2\angle C = 90^{\circ}$. 设$CE$与$BD$的交点为$G$.$\because\angle D+\angle DAE+\angle DGA = 180^{\circ},\angle DGA+\angle CGB = 180^{\circ},\therefore\angle CGB=\angle D+\angle DAE$.$\because BD\perp BC,\therefore\angle CBD = 90^{\circ}$. 在三角形$BCG$中，$\angle CBD+\angle CGB+\angle C = 180^{\circ}.\therefore\angle CGB+\angle C = 90^{\circ}.\therefore\angle D+\angle DAE+\angle C = 90^{\circ}$. 又$\because\angle D=\angle C,\therefore 2\angle C+\angle DAE = 90^{\circ}$.
(3)设$\angle DAE=\alpha$，则$\angle DFE = 8\alpha$.$\because\angle DFE+\angle AFD = 180^{\circ},\therefore\angle AFD = 180^{\circ}-8\alpha$.$\because DF// BC,\therefore\angle C=\angle AFD = 180^{\circ}-8\alpha$. 又$\because 2\angle C+\angle DAE = 90^{\circ}$，$\therefore 2(180^{\circ}-8\alpha)+\alpha = 90^{\circ}.\therefore\alpha = 18^{\circ}$.$\therefore\angle C = 180^{\circ}-8\alpha = 36^{\circ}=\angle ADB$. 又$\because\angle C=\angle BDA,\angle BAC=\angle BAD$，$\therefore\angle ABC=\angle ABD=\frac{1}{2}\angle CBD = 45^{\circ}$.$\therefore$在三角形$ABD$中，$\angle BAD = 180^{\circ}-45^{\circ}-36^{\circ}=99^{\circ}$.