《2025年假期面对面南方出版社八年级数学沪科版》

12.(黑龙江中考)如图,$Rt\triangle ABC$和$Rt\triangle EDF$中,$∠B=∠D$,在不添加任何辅助线的情况下,请你添加一个条件______$ AB = ED $（答案不唯一）,使$Rt\triangle ABC$和$Rt\triangle EDF$全等.

答案： $ AB = ED $（答案不唯一）

13.(成都中考)如图,在$\triangle ABC$中,$AB=AC$,点D,E都在边BC上,$∠BAD=∠CAE$,若$BD=9$,则CE的长为______9.

答案： 9

14.(8分)(铜仁中考)如图,$∠B=∠E,BF=EC,AC// DF$.求证:$\triangle ABC\cong \triangle DEF$.
证明：$\because AC // DF$，$\therefore ∠ACB = ∠DFE$。$\because BF = EC$，$\therefore BC = EF$。在$\triangle ABC$和$\triangle DEF$中，$\because \left\{ \begin{array} { l } { ∠B = ∠E, } \\ { BC = EF, } \\ { ∠ACB = ∠DFE, } \end{array} \right.$ $\therefore \triangle ABC \cong \triangle DEF (ASA)$。

答案： 证明：$\because AC // DF$，$\therefore \angle ACB = \angle DFE$。$\because BF = EC$，$\therefore BC = EF$。在$\triangle ABC$和$\triangle DEF$中，$\because \left\{ \begin{array} { l } { B = E, } \\ { B C = E F, } \\ { \angle A C B = \angle D F E, } \end{array} \right.$ $\therefore \triangle ABC \cong \triangle DEF ( \text { ASA } )$。

15.(8分)(南充中考)如图,点C在线段BD上,且$AB⊥BD,DE⊥BD,AC⊥CE,BC=DE$.求证:$AB=CD$.
证明：$\because AB \perp BD$，$ED \perp BD$，$AC \perp CE$，$\therefore \angle ACE = \angle ABC = \angle CDE = 90 ^ { \circ }$。$\therefore \angle ACB + \angle ECD = 90 ^ { \circ }$，$\angle ECD + \angle CED = 90 ^ { \circ }$，$\therefore \angle ACB = \angle CED$。在$\triangle ABC$和$\triangle CDE$中，$\because \left\{ \begin{array} { l } { \angle A C B = \angle C E D, } \\ { B C = D E, } \\ { \angle A B C = \angle C D E, } \end{array} \right.$ $\therefore \triangle ABC \cong \triangle CDE ( \text { ASA } )$。$\therefore AB = CD$。

答案： 证明：$\because AB \perp BD$，$ED \perp BD$，$AC \perp CE$，$\therefore \angle ACE = \angle ABC = \angle CDE = 90 ^ { \circ }$。$\therefore \angle ACB + \angle ECD = 90 ^ { \circ }$，$\angle ECD + \angle CED = 90 ^ { \circ }$，$\therefore \angle ACB = \angle CED$。在$\triangle ABC$和$\triangle CDE$中，$\because \left\{ \begin{array} { l } { \angle A C B = \angle C E D, } \\ { B C = D E, } \\ { \angle A B C = \angle C D E, } \end{array} \right.$ $\therefore \triangle ABC \cong \triangle CDE ( \text { ASA } )$。$\therefore AB = CD$。

16.(16分)(常州中考)已知:如图,点A,B,C,D在一条直线上,$EA// FB,EA=FB,AB=CD$.
(1)求证:$∠E=∠F$;
证明：$\because EA // FB$，$\therefore \angle A = \angle FBD$。$\because AB = CD$，$\therefore AB + BC = CD + BC$，即$AC = BD$。在$\triangle EAC$和$\triangle FBD$中，$\because \left\{ \begin{array} { l } { E A = F B, } \\ { \angle A = \angle F B D, } \\ { A C = B D, } \end{array} \right.$ $\therefore \triangle EAC \cong \triangle FBD ( \text { SAS } )$。$\therefore \angle E = \angle F$。
(2)若$∠A=40^{\circ },∠D=80^{\circ }$,求$∠E$的度数.
解：$\because \triangle EAC \cong \triangle FBD$，$\therefore \angle ECA = \angle D = 80 ^ { \circ }$。$\because \angle A = 40 ^ { \circ }$，$\therefore \angle E = 180 ^ { \circ } - 40 ^ { \circ } - 80 ^ { \circ } = 60 ^ { \circ }$。
故$∠E$的度数为60°。

答案：
(1) 证明：$\because EA // FB$，$\therefore \angle A = \angle FBD$。$\because AB = CD$，$\therefore AB + BC = CD + BC$，即$AC = BD$。在$\triangle EAC$和$\triangle FBD$中，$\because \left\{ \begin{array} { l } { E A = F B, } \\ { \angle A = \angle F B D, } \\ { A C = B D, } \end{array} \right.$ $\therefore \triangle EAC \cong \triangle FBD ( \text { SAS } )$。$\therefore \angle E = \angle F$。
(2) 解：$\because \triangle EAC \cong \triangle FBD$，$\therefore \angle ECA = \angle D = 80 ^ { \circ }$。$\because \angle A = 40 ^ { \circ }$，$\therefore \angle E = 180 ^ { \circ } - 40 ^ { \circ } - 80 ^ { \circ } = 60 ^ { \circ }$。

17.(16分)(宜宾中考)如图,在$\triangle ABC$中,点D是边BC的中点,连接AD并延长到点E,使$DE=AD$,连接CE.
(1)求证:$\triangle ABD\cong \triangle ECD$;
证明：$\because D$是$BC$的中点，$\therefore BD = CD$。在$\triangle ABD$和$\triangle ECD$中，$\because \left\{ \begin{array} { l } { B D = C D, } \\ { \angle A D B = \angle E D C, } \\ { A D = E D, } \end{array} \right.$ $\therefore \triangle ABD \cong \triangle ECD ( \text { SAS } )$。
(2)若$\triangle ABD$的面积为5,求$\triangle ACE$的面积.
解：在$\triangle ABC$中，$\because D$是边$BC$的中点，$\therefore S _ { \triangle ABD } = S _ { \triangle ACD }$。$\because \triangle ABD \cong \triangle ECD$，$\therefore S _ { \triangle ABD } = S _ { \triangle ECD }$。$\because S _ { \triangle ABD } = 5$，$\therefore S _ { \triangle ACE } = S _ { \triangle ACD } + S _ { \triangle ECD } = 5 + 5 = 10$。

答案：
(1) 证明：$\because D$是$BC$的中点，$\therefore BD = CD$。在$\triangle ABD$和$\triangle ECD$中，$\because \left\{ \begin{array} { l } { B D = C D, } \\ { \angle A D B = \angle E D C, } \\ { A D = E D, } \end{array} \right.$ $\therefore \triangle ABD \cong \triangle ECD ( \text { SAS } )$。
(2) 解：在$\triangle ABC$中，$\because D$是边$BC$的中点，$\therefore S _ { \triangle ABD } = S _ { \triangle ACD }$。$\because \triangle ABD \cong \triangle ECD$，$\therefore S _ { \triangle ABD } = S _ { \triangle ECD }$。$\because S _ { \triangle ABD } = 5$，$\therefore S _ { \triangle ACE } = S _ { \triangle ACD } + S _ { \triangle ECD } = 5 + 5 = 10$。