《2025年暑假总动员八年级数学沪科版合肥工业大学出版社》

9.(衡阳)如图,点A,B,D,E在同一条直线上,$AB= DE,AC// DF,BC// EF$.求证:$△ABC\cong △DEF$.
证明：$ \because A C // D F $，$ \therefore $$\angle C A B = \angle F D E$(两直线平行，同位角相等)。又$ \because B C // E F $，$ \therefore $$\angle C B A = \angle F E D$(两直线平行，同位角相等)。在$ \triangle A B C $和$ \triangle D E F $中，$ \left\{ \begin{array} { l } { \angle C A B = \angle F D E }, \\ { A B = D E }, \\ { \angle C B A = \angle F E D }, \end{array} \right. $ $ \therefore \triangle A B C \cong \triangle D E F $(ASA)。

答案： 证明：$ \because A C // D F $，$ \therefore \angle C A B = \angle F D E $(两直线平行，同位角相等)。又$ \because B C // E F $，$ \therefore \angle C B A = \angle F E D $(两直线平行，同位角相等)。在$ \triangle A B C $和$ \triangle D E F $中，$ \left\{ \begin{array} { l } { \angle C A B = \angle F D E }, \\ { A B = D E }, \\ { \angle C B A = \angle F E D }, \end{array} \right. $ $ \therefore \triangle A B C \cong \triangle D E F ( A S A ) $。

10.如图,$AC⊥BC,DC⊥EC,AC= BC,DC= EC$,AE与BD交于点F.
(1)求证:$AE=$BD;
(2)求$∠AFD$的度数.90°

答案：
(1) 证明：$ \because A C \perp B C $，$ D C \perp E C $，$ \therefore \angle A C B = \angle D C E = 90 ^ { \circ } $，$ \therefore \angle A C E = \angle B C D $。在$ \triangle A C E $和$ \triangle B C D $中，$ \left\{ \begin{array} { l } { A C = B C }, \\ { \angle A C E = \angle B C D }, \\ { C E = C D }, \end{array} \right. $ $ \therefore \triangle A C E \cong \triangle B C D ( S A S ) $，$ \therefore A E = B D $。
(2) 解：设$ B C $与$ A E $交于点$ N $，$ \because \angle A C B = 90 ^ { \circ } $，$ \therefore \angle A + \angle A N C = 90 ^ { \circ } $。$ \because \triangle A C E \cong \triangle B C D $，$ \therefore \angle A = \angle B $。$ \because \angle A N C = \angle B N F $，$ \therefore \angle B + \angle B N F = \angle A + \angle A N C = 90 ^ { \circ } $，$ \therefore \angle A F D = \angle B + \angle B N F = 90 ^ { \circ } $。

11.(大连)如图,点A,D,B,E在一条直线上,$AD= BE,AC= DF,AC// DF$.求证:$BC= EF$.
证明：$ \because A D = B E $，$ \therefore A D + B D = B E + B D $，即$ A B = $DE。$ \because A C // D F $，$ \therefore \angle A = $∠EDF。在$ \triangle A B C $与$ \triangle D E F $中，$ \left\{ \begin{array} { l } { A B = D E }, \\ { \angle A = \angle E D F }, \\ { A C = D F }, \end{array} \right. $ $ \therefore \triangle A B C \cong \triangle D E F $(SAS)，$ \therefore B C = E F $。

答案： 证明：$ \because A D = B E $，$ \therefore A D + B D = B E + B D $，即$ A B = D E $。$ \because A C // D F $，$ \therefore \angle A = \angle E D F $。在$ \triangle A B C $与$ \triangle D E F $中，$ \left\{ \begin{array} { l } { A B = D E }, \\ { \angle A = \angle E D F }, \\ { A C = D F }, \end{array} \right. $ $ \therefore \triangle A B C \cong \triangle D E F ( S A S ) $，$ \therefore B C = E F $。

12.(常州)如图,B,F,C,E是直线l上的四点,$AB// DE,AB= DE,BF= CE$.
(1)求证:$△ABC\cong △DEF;$
(2)将$△ABC$沿直线l翻折得到$△A'BC$.
①用直尺和圆规在图中作出$△A'BC$(保留作图痕迹,不要求写作法);
②连接$A'D$,则直线$A'D$与l的位置关系是______.

答案：

(1) 证明：$ \because B F = C E $，$ \therefore B F + F C = C E + F C $，即$ B C = E F $。$ \because A B // D E $，$ \therefore \angle A B C = \angle D E F $，在$ \triangle A B C $与$ \triangle D E F $中，$ \left\{ \begin{array} { l } { A B = D E }, \\ { \angle A B C = \angle D E F }, \\ { B C = E F }, \end{array} \right. $ $ \therefore \triangle A B C \cong \triangle D E F ( S A S ) $。
(2) 解：①如图所示，$ \triangle A ^ { \prime } B C $即为所求。
②$ A ^ { \prime } D // l $

13.(黄石)如图,D是$△ABC$的边AB上一点,$CF// AB$,DF交AC于E点,$DE= EF$.
(1)求证:$△ADE\cong △CFE;$
证明：$ \because F C // A B $，$ \therefore \angle A = \angle A C F $，在$ \triangle A D E $和$ \triangle C F E $中，$ \left\{ \begin{array} { l } { \angle A = \angle E C F }, \\ { \angle A E D = \angle C E F }, \\ { D E = E F }, \end{array} \right. $ $ \therefore \triangle A D E \cong \triangle C F E $(AAS)。
(2)若$AB= 5,CF= 4$,求BD的长.
解：$ \because A B = 5 $，$ C F = 4 $，由(1)知$ A D = C F = 4 $，$ \therefore B D = A B - A D = 5 - 4 = $1。

答案：
(1) 证明：$ \because F C // A B $，$ \therefore \angle A = \angle A C F $，在$ \triangle A D E $和$ \triangle C F E $中，$ \left\{ \begin{array} { l } { \angle A = \angle E C F }, \\ { \angle A E D = \angle C E F }, \\ { D E = E F }, \end{array} \right. $ $ \therefore \triangle A D E \cong \triangle C F E ( A A S ) $。
(2) 解：$ \because A B = 5 $，$ C F = 4 $，由
(1)知$ A D = C F = 4 $，$ \therefore B D = A B - A D = 5 - 4 = 1 $。

14.(南充)如图,$∠BAC= 90^{\circ }$,AD是$∠BAC$内部一条射线,若$AB= AC,BE⊥AD$于点E,$CF⊥AD$于点F.求证:$AF= BE$.
证明：$ \because \angle B A C = 90 ^ { \circ } $，$ \therefore \angle B A E + \angle F A C = 90 ^ { \circ } $。$ \because B E \perp A D $，$ C F \perp A D $，$ \therefore \angle B E A = \angle A F C = 90 ^ { \circ } $，$ \therefore \angle B A E + \angle E B A = 90 ^ { \circ } $，$ \therefore \angle E B A = \angle F A C $，在$ \triangle A C F $和$ \triangle B A E $中，$ \left\{ \begin{array} { l } { \angle A F C = \angle B E A }, \\ { \angle F A C = \angle E B A }, \\ { A C = B A }, \end{array} \right. $ $ \therefore \triangle A C F \cong \triangle B A E ( A A S ) $，$ \therefore A F = B E $。

答案： 证明：$ \because \angle B A C = 90 ^ { \circ } $，$ \therefore \angle B A E + \angle F A C = 90 ^ { \circ } $。$ \because B E \perp A D $，$ C F \perp A D $，$ \therefore \angle B E A = \angle A F C = 90 ^ { \circ } $，$ \therefore \angle B A E + \angle E B A = 90 ^ { \circ } $，$ \therefore \angle E B A = \angle F A C $，在$ \triangle A C F $和$ \triangle B A E $中，$ \left\{ \begin{array} { l } { \angle A F C = \angle B E A }, \\ { \angle F A C = \angle E B A }, \\ { A C = B A }, \end{array} \right. $ $ \therefore \triangle A C F \cong \triangle B A E ( A A S ) $，$ \therefore A F = B E $。