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1. (2025·青海中考)工人师傅常用角尺平分一个任意角.做法如下:如图,$∠ AOB$是一个任意角,在边$OA$,$OB$上分别取$OM=ON$,移动角尺,使角尺两边相同的刻度分别与点$M$,$N$重合,即$CM=CN$,过角尺顶点$C$的射线$OC$便是$∠ AOB$的平分线,这种做法的依据是 （
A.AAS
B.SAS
C.SSS
D.ASA
C
）A.AAS
B.SAS
C.SSS
D.ASA
答案
1. C 解析:在$△ ONC$和$△ OMC$中,$\begin{cases} MO = NO,\\ MC = NC,\\ OC = OC, \end{cases}$$\therefore △ MOC ≌ △ NOC(\mathrm{SSS})$,$\therefore ∠ BOC = ∠ AOC$,故选C.
2.(扬州中考)如图,小明家仿古家具的一块三角形形状的玻璃坏了,需要重新配一块.小明通过电话给玻璃店老板提供相关数据,为了方便表述,将该三角形记为$△ ABC$,提供了下列各组元素的数据,配出来的玻璃不一定符合要求的是（
A.$AB$,$BC$,$CA$
B.$AB$,$BC$,$∠ B$
C.$AB$,$AC$,$∠ B$
D.$∠ A$,$∠ B$,$BC$
C
）A.$AB$,$BC$,$CA$
B.$AB$,$BC$,$∠ B$
C.$AB$,$AC$,$∠ B$
D.$∠ A$,$∠ B$,$BC$
答案
2. C 解析:A.利用SSS判定两三角形全等,三角形形状确定,故此选项不合题意;B.利用SAS判定两三角形全等,三角形形状确定,故此选项不合题意;C.$AB,AC,∠ B$,无法确定三角形的形状,故此选项符合题意;D.根据AAS判定两三角形全等,三角形形状确定,故此选项不合题意.故选C.
3. (2026·南通校级月考) 如图,在$△ ABC$中,点$D,E$分别在$AB,BC$上,且$∠ A + ∠ DEB = 180°$,请你添加一个条件,使$△ ADC ≌ △ EDC$.你所添加的条件是
$∠ ECD = ∠ ACD$(答案不唯一)
(添加一个即可).答案
3. $∠ ECD = ∠ ACD$(答案不唯一) 解析:添加的条件是:$∠ ECD = ∠ ACD$,理由是:$\because ∠ A + ∠ DEB = 180°$,$∠ DEC + ∠ DEB = 180°$,$\therefore ∠ A = ∠ DEC$.在$△ CDE$和$△ CDA$中,$\begin{cases} ∠ DEC = ∠ A,\\ ∠ ECD = ∠ ACD,\\ CD = CD, \end{cases}$$\therefore △ CDE ≌ △ CDA(\mathrm{AAS})$.
4. (2025·宜宾期中)如图是由边长相同的小正方形组成的网格,A,B,C,D,E五点均在格点上,则$∠ ABC+∠ ADE$的度数为
$180°$
.答案
4. $180°$ 解析:如图,在$△ ABF$与$△ ADE$中,$\begin{cases} AF = AE,\\ ∠ F = ∠ E,\\ FB = ED, \end{cases}$$\therefore △ ABF ≌ △ ADE\ (\mathrm{SAS})$,$\therefore ∠ ABF = ∠ ADE$.$\because ∠ ABC + ∠ ABF = 180°$,$\therefore ∠ ABC + ∠ ADE = 180°$.
5. 如图,在多边形$ABCDE$中,$BC ⊥ CD$,$BF ⊥ AE$于点$F$,且$BF=BC$,$∠ CBF=2∠ DBE$,$∠ ABF=∠ CBD$.
(1)求证:$AB=DB$;
(2)若$DE=4$,$BF=3$,求$△ BDE$的面积.
(1)求证:$AB=DB$;
(2)若$DE=4$,$BF=3$,求$△ BDE$的面积.
答案
5. (1)$\because BC ⊥ CD$,$BF ⊥ AE$,$\therefore ∠ AFB = ∠ C = 90°$.在$△ BFA$和$△ BCD$中,$\begin{cases} ∠ AFB = ∠ C = 90°,\\ BF = BC,\\ ∠ ABF = ∠ CBD, \end{cases}$$\therefore △ BFA ≌ △ BCD\ (\mathrm{ASA})$,$\therefore AB = DB$.
(2)$\because ∠ CBF = 2∠ DBE$,$\therefore ∠ CBD + ∠ DBE + ∠ EBF = 2∠ DBE$,即$∠ CBD + ∠ EBF = ∠ DBE$.$\because ∠ ABF = ∠ CBD$,$\therefore ∠ ABF + ∠ EBF = ∠ DBE$,即$∠ ABE = ∠ DBE$.在$△ ABE$和$△ DBE$中,$\begin{cases} AB = BD,\\ ∠ ABE = ∠ DBE,\\ BE = BE, \end{cases}$$\therefore S_{△ ABE} = S_{△ DBE}$,$AE = DE$.$\because DE = 4$,$\therefore AE = DE = 4$.$\because BF ⊥ AE$,$\therefore S_{△ ABE} = \dfrac{1}{2}AE · BF = \dfrac{1}{2} × 4 × 3 = 6$,$\therefore S_{△ DBE} = 6$.
(2)$\because ∠ CBF = 2∠ DBE$,$\therefore ∠ CBD + ∠ DBE + ∠ EBF = 2∠ DBE$,即$∠ CBD + ∠ EBF = ∠ DBE$.$\because ∠ ABF = ∠ CBD$,$\therefore ∠ ABF + ∠ EBF = ∠ DBE$,即$∠ ABE = ∠ DBE$.在$△ ABE$和$△ DBE$中,$\begin{cases} AB = BD,\\ ∠ ABE = ∠ DBE,\\ BE = BE, \end{cases}$$\therefore S_{△ ABE} = S_{△ DBE}$,$AE = DE$.$\because DE = 4$,$\therefore AE = DE = 4$.$\because BF ⊥ AE$,$\therefore S_{△ ABE} = \dfrac{1}{2}AE · BF = \dfrac{1}{2} × 4 × 3 = 6$,$\therefore S_{△ DBE} = 6$.
6. (2026·潍坊期末)山东潍坊是世界风筝都，匠人在制作过程中采用了全等的相关知识.在如图所示的风筝“龙骨”图案中，$AB = AD$，$∠ B=∠ D,BC=DE$.则不一定能得到以下哪个结论
（
A.$△ ABC ≌ △ ADE$
B.$△ ABF ≌ △ ADG$
C.$FC = GE$
D.$AG = GE$
（
D
）A.$△ ABC ≌ △ ADE$
B.$△ ABF ≌ △ ADG$
C.$FC = GE$
D.$AG = GE$
答案
6. D 解析:在$△ ABC$和$△ ADE$中,$\begin{cases} AB = AD,\\ ∠ B = ∠ D,\\ BC = DE, \end{cases}$$\therefore △ ABC ≌ △ ADE(\mathrm{SAS})$,故选项A不符合题意;$\therefore ∠ BAC = ∠ DAE$,$BC = DE$,$\therefore ∠ BAC - ∠ EAC = ∠ DAE - ∠ EAC$,即$∠ BAE = ∠ DAG$.$\because AB = AD$,$∠ B = ∠ D$,$\therefore △ ABF ≌ △ ADG(\mathrm{ASA})$,故选项B不符合题意;$\therefore BF = DG$,$\therefore BC - BF = DE - DG$,即$FC = GE$,故选项C不符合题意;无法证明$AG = GE$,故选项D符合题意.故选D.
7. (2025·太原模拟)如图,已知$△ ABC$,点$D$是$AC$边上一点,根据尺规作图的痕迹,能确定线段$BD$是$△ ABC$的（
A.中线
B.角平分线
C.高线
D.以上均正确
B
）A.中线
B.角平分线
C.高线
D.以上均正确
答案
7. B 解析:如图,根据作图可知,$BE = BF$,$BG = BH$,又$\because ∠ EBG = ∠ FBH$,$\therefore △ BEG ≌ △ BFH(\mathrm{SAS})$,$\therefore ∠ EHM = ∠ FGM$.$\because BH - BE = BG - BF$,$\therefore EH = FG$.又$\because ∠ EMH = ∠ FMG$,$\therefore △ EMH ≌ △ FMG(\mathrm{AAS})$,$\therefore GM = HM$.又$\because BG = BH$,$BM = BM$,$\therefore △ BMH ≌ △ BMG(\mathrm{SSS})$,$\therefore ∠ HBM = ∠ GBM$,$\therefore BD$平分$∠ ABC$,即线段$BD$是$△ ABC$的角平分线.故选B.
