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1. (2024·射阳县月考)如图，$∠ B=∠ D$，$DE=BC$，若$AB=8$，$AC=3$，则$DC$的长是 （
A.5
B.4
C.3
D.5.5
A
）A.5
B.4
C.3
D.5.5
答案
1.A
2. (2024·大丰区月考)如图，$BD=BC,BE=CA,∠ DBE=∠ C=62^{\circ },∠ BDE=75^{\circ }$,则$∠ AFD$的度数等于（
A.$30^{\circ }$
B.$32^{\circ }$
C.$33^{\circ }$
D.$35^{\circ }$
B
）A.$30^{\circ }$
B.$32^{\circ }$
C.$33^{\circ }$
D.$35^{\circ }$
答案
2.B
3.(2024·阜宁县月考)如图,在$2×2$的正方形网格中,$∠ 1+∠ 2=$
90
$°$.答案
3.90
4.(2024·盐都区期中)如图,小明用“X”型转动钳测量圆柱形小口容器壁的厚度.已知$OA=$$OD,OB=OC,AB=6\ \mathrm{cm},EF=8\ \mathrm{cm}$,则该容器壁的厚度为
1
cm.答案
4.1
5. 如图, 在$△ ABC$和$△ ADE$中, 延长$BC$交$DE$于点$F$. 已知$BC = DE$,$∠ BAD = ∠ CAE$,$∠ ACF + ∠ AED = 180°$. 求证:$AB = AD$.
答案
5. 证明:$\because ∠BAD=∠CAE$,
$\therefore ∠BAD-∠CAD=∠CAE-∠CAD$,即$∠BAC=∠DAE$.
$\because ∠ACF+∠AED=180°=∠ACF+∠ACB$,
$\therefore ∠AED=∠ACB$.
在$△ ABC$和$△ ADE$中,$\begin{cases} ∠BAC=∠DAE, \\ ∠ACB=∠AED, \\ BC=DE, \end{cases}$
$\therefore △ ABC≌△ ADE(\mathrm{AAS}),\therefore AB=AD.$
$\therefore ∠BAD-∠CAD=∠CAE-∠CAD$,即$∠BAC=∠DAE$.
$\because ∠ACF+∠AED=180°=∠ACF+∠ACB$,
$\therefore ∠AED=∠ACB$.
在$△ ABC$和$△ ADE$中,$\begin{cases} ∠BAC=∠DAE, \\ ∠ACB=∠AED, \\ BC=DE, \end{cases}$
$\therefore △ ABC≌△ ADE(\mathrm{AAS}),\therefore AB=AD.$
6. (2024·亭湖区期中)如图，$AB ⊥ BC$，$BE ⊥ AC$，垂足分别为$B$，$E$，$∠ 1 = ∠ 2$，$AD = AB$，则下列结论正确的是（
A.$∠ 1 = ∠ EFD$
B.$BE = EC$
C.$BF = CD$
D.$FD // BC$
D
）A.$∠ 1 = ∠ EFD$
B.$BE = EC$
C.$BF = CD$
D.$FD // BC$
答案
6.D
