8.(2025·厦门期中)如图,在$△ABC和△BDE$中,点C在边BD上,边AC交边BE于点F.若$AC= BD,AB= ED,BC= BE$,则$∠ACB$等于 ()

A.$∠EDB$
B.$∠BED$
C.$2∠ABF$
D.$\frac {1}{2}∠AFB$

D 解析：在$\triangle ABC$和$\triangle DEB$中，$\left\{\begin{array}{l} AC = DB,\\ AB = DE,\\ BC = EB,\end{array}\right. $ $\therefore \triangle ABC≌\triangle DEB(SSS)$，$\therefore ∠ACB = ∠DBE$。$\because ∠AFB$是$\triangle BFC$的外角，$\therefore ∠ACB + ∠DBE = ∠AFB$，$\therefore ∠ACB = \frac{1}{2}∠AFB$，故选D。

9.如图,$CA= CB,AD= BD$,M,N分别是CA,CB的中点,若$△ADM的面积为\frac {3}{2}$,则图中阴影部分的面积为____.

3 解析：连接$CD$，$\because M$，$N$分别是$CA$，$CB$的中点，$\therefore S_{\triangle CDM} = S_{\triangle ADM} = \frac{1}{2}S_{\triangle ACD} = \frac{3}{2}$，$S_{\triangle CDN} = S_{\triangle BDN} = \frac{1}{2}S_{\triangle BCD}$，$\therefore S_{\triangle ACD} = 3$。在$\triangle ACD$和$\triangle BCD$中，$\left\{\begin{array}{l} CA = CB,\\ AD = BD,\\ CD = CD,\end{array}\right. $ $\therefore \triangle ACD≌\triangle BCD(SSS)$，$\therefore S_{\triangle BCD} = S_{\triangle ACD} = 3$，$\therefore S_{\triangle CDN} = \frac{1}{2}S_{\triangle BCD} = \frac{3}{2}$，$\therefore S_{阴影} = S_{\triangle CDM} + S_{\triangle CDN} = 3$。

10.(2025·德州期中)如图,勤劳的小蜜蜂A,B,C,D,E,F分别位于蜂房(由若干个正六边形拼成)向阳面的一侧劳作,若任何不共线三点位置都可以组成一个三角形,则与$△ACD$全等的三角形是____.

$\triangle ABC$，$\triangle ADE$ 解析：如图，根据图形可知$AD = AD$，$AE = AC$，$DE = DC$，在$\triangle ACD$和$\triangle AED$中，$\left\{\begin{array}{l} AD = AD,\\ AC = AE,\\ DC = DE,\end{array}\right. $ $\therefore \triangle ACD≌\triangle AED(SSS)$。根据图形可知$AC = AC$，$AD = BC$，$AB = DC$，在$\triangle ACD$和$\triangle CAB$中，$\left\{\begin{array}{l} AC = CA,\\ AD = CB,\\ DC = BA,\end{array}\right. $ $\therefore \triangle ACD≌\triangle CAB(SSS)$。

11.如图,在四边形ABCD中,$AB= AD,∠B+∠ADC= 180^{\circ }$,点E,F分别是BC,CD上的点,且$EF= BE+FD$,连接AE,AF.延长FD到点G,使$DG= BE$,连接AG.若$∠EAF= 55^{\circ }$,则$∠FAG$的度数为____$^{\circ }$.

55 解析：$\because ∠B + ∠ADC = 180^{\circ}$，$∠ADG + ∠ADC = 180^{\circ}$，$\therefore ∠B = ∠ADG$。在$\triangle ABE$与$\triangle ADG$中，$\left\{\begin{array}{l} AB = AD,\\ ∠B = ∠ADG,\\ BE = DG,\end{array}\right. $ $\therefore \triangle ABE≌\triangle ADG(SAS)$，$\therefore AE = AG$，$∠BAE = ∠DAG$。$\because EF = BE + DF$，$FG = DG + DF = BE + DF$，$\therefore EF = FG$。在$\triangle AEF$与$\triangle AGF$中，$\left\{\begin{array}{l} AE = AG,\\ AF = AF,\\ EF = GF,\end{array}\right. $ $\therefore \triangle AEF≌\triangle AGF(SSS)$，$\therefore ∠EAF = ∠GAF = 55^{\circ}$。

12.如图,在四边形ABCD中,$CB⊥AB$于点B,$CD⊥AD$于点D,点E,F分别在AB,AD上,$AE= AF,CE= CF.$
(1)求证:$CB= CD;$
(2)若$AE= 8,CD= 6$,求四边形AECF的面积;
(3)猜想$∠DAB,∠ECF,∠DFC$三者之间的数量关系,并证明你的猜想.

(1) 连接$AC$，在$\triangle ACE$和$\triangle ACF$中，$\left\{\begin{array}{l} AE = AF,\\ CE = CF,\\ AC = AC,\end{array}\right. $ $\therefore \triangle ACE≌\triangle ACF(SSS)$，$\therefore ∠FAC = ∠EAC$。$\because CB⊥AB$，$CD⊥AD$，$\therefore ∠B = ∠D = 90^{\circ}$。$\because AC = AC$，$\therefore \triangle ACB≌\triangle ACD(AAS)$。$\therefore CB = CD$。
(2) 由(1)得$\triangle ACE≌\triangle ACF$，$CB = CD$。$\because AE = 8$，$CD = 6$，$\therefore S_{\triangle ACF} = S_{\triangle ACE} = \frac{1}{2}AE\cdot CB = \frac{1}{2}×8×6 = 24$，$\therefore S_{四边形AECF} = S_{\triangle ACF} + S_{\triangle ACE} = 24 + 24 = 48$。
(3)$∠DAB + ∠ECF = 2∠DFC$。证明：$\because \triangle ACE≌\triangle ACF$，$\therefore ∠EAC = ∠FAC$，$∠ACE = ∠ACF$。$\because ∠DAB = ∠FAC + ∠EAC$，$∠ECF = ∠ACF + ∠ACE$，$\therefore ∠DAB + ∠ECF = ∠FAC + ∠EAC + ∠ACF + ∠ACE = 2∠FAC + 2∠ACF = 2(∠FAC + ∠ACF)$。$\because ∠DFC = ∠FAC + ∠ACF$，$\therefore ∠DAB + ∠ECF = 2∠DFC$。

13.已知$△ABC$中,$AB= BC≠AC$,作与$△ABC$只有一条公共边,且与$△ABC$全等的三角形,这样的三角形一共能作出____个.

7 解析：只要满足三边对应相等(SSS)就能保证作出的三角形与原三角形全等，如图，以腰为公共边时有6个，以底为公共边时有1个。

14.如图,$AD= CB$,E,F是AC上两动点,且$DE= BF.$
(1)若点E,F运动至如图①所示的位置,且有$AF= CE$,求证:$△ADE\cong △CBF.$
(2)若点E,F运动至如图②所示的位置,仍有$AF= CE$,则$△ADE\cong △CBF$还成立吗? 为什么?
(3)若点E,F不重合,则AD和CB平行吗?请说明理由.

(1)$\because AF = CE$，$\therefore AF + EF = CE + EF$，即$AE = CF$。在$\triangle ADE$和$\triangle CBF$中，$\left\{\begin{array}{l} AD = CB,\\ DE = BF,\\ AE = CF,\end{array}\right. $ $\therefore \triangle ADE≌\triangle CBF(SSS)$。
(2)$\triangle ADE≌\triangle CBF$成立。理由如下：$\because AF = CE$，$\therefore AF - EF = CE - EF$，即$AE = CF$。在$\triangle ADE$和$\triangle CBF$中，$\left\{\begin{array}{l} AD = CB,\\ DE = BF,\\ AE = CF,\end{array}\right. $ $\therefore \triangle ADE≌\triangle CBF(SSS)$。
(3)$AD$与$CB$不一定平行。理由如下：在$\triangle ADE$和$\triangle CBF$中，仅有$AD = CB$，$DE = BF$，不能判定它们全等，即不能得出$∠A = ∠C$，故$AD$与$CB$不一定平行。