《2025年轻松作业本八年级数学上册苏科版》

1. 如图,在四边形ABCD中,$∠BCD= 90^{\circ },AB⊥BD$于点B,点E是BD的中点,连接AE、CE,则AE与CE的大小关系是(C)

A. $AE＜CE$
B. $AE= CE$
C. $AE＞CE$
D. $AE= 2CE$

答案： C

2. 如图,一架梯子AB斜靠在竖直墙上,点M为梯子AB的中点,当梯子底端向左水平滑动到CD位置时,滑动过程中OM的变化规律是(B)

A. 变小
B. 不变
C. 变大
D. 先变小再变大

答案： B

3. 如图,在$Rt△ABC$中,CD是斜边AB上的中线,$∠A= 20^{\circ }$,则$∠BCD$的度数是$70^{\circ}$．

答案： $70^{\circ}$

4. 如图,$△ABC$中,$AB= 5,AC= 8$,BD、CD分别平分$∠ABC$、$∠ACB$,过点D作直线$EF// BC$,分别交AB、AC于E、F,则$△AEF$的周长为______13．

答案： 13

5. 如图,在$△ABC$中,$AB= AC$,D为BC边上一点,$∠B= 30^{\circ },∠DAB= 45^{\circ }$．
(1)求$∠DAC$的度数；
解：$\because AB = AC$，$\therefore \angle B = \angle C = 30^{\circ}$。$\because \angle C + \angle BAC + \angle B = 180^{\circ}$，$\therefore \angle BAC = 180^{\circ} - 30^{\circ} - 30^{\circ} = 120^{\circ}$。$\because \angle DAB = 45^{\circ}$，$\therefore \angle DAC = \angle BAC - \angle DAB = 120^{\circ} - 45^{\circ} =$75°；
(2)求证：$DC= AB$．
证明：$\because \angle DAB = 45^{\circ}$，$\therefore \angle ADC = \angle B + \angle DAB =$75°。$\therefore \angle DAC = \angle ADC$。$\therefore DC =$AC。$\therefore DC = AB$。

答案： （1）解：$\because AB = AC$，$\therefore \angle B = \angle C = 30^{\circ}$。$\because \angle C + \angle BAC + \angle B = 180^{\circ}$，$\therefore \angle BAC = 180^{\circ} - 30^{\circ} - 30^{\circ} = 120^{\circ}$。$\because \angle DAB = 45^{\circ}$，$\therefore \angle DAC = \angle BAC - \angle DAB = 120^{\circ} - 45^{\circ} = 75^{\circ}$；
（2）证明：$\because \angle DAB = 45^{\circ}$，$\therefore \angle ADC = \angle B + \angle DAB = 75^{\circ}$。$\therefore \angle DAC = \angle ADC$。$\therefore DC = AC$。$\therefore DC = AB$。

6. 如图,在$△ABC$中,$∠B= 90^{\circ },AB= BC,BD= CE$,M是AC边上的中点．求证：$△DEM$是等腰三角形．
证明：连接BM，$\because \angle B = 90^{\circ}$，$AB =$BC，$\therefore BM = \frac{1}{2}AC = AM = MC$，$\angle A = \angle C = 45^{\circ}$，$\angle ABM = \angle CBM = 45^{\circ}$。又 $\because BD = CE$，$\therefore \triangle DBM \cong \triangle ECM$(SAS)。$\therefore$DM = EM。$\therefore \triangle DEM$ 是等腰三角形。

答案： 证明：连接 $BM$，$\because \angle B = 90^{\circ}$，$AB = AC$，$\therefore BM = \frac{1}{2}AC = AM = MC$，$\angle A = \angle C = 45^{\circ}$，$\angle ABM = \angle CBM = 45^{\circ}$。又 $\because BD = CE$，$\therefore \triangle DBM \cong \triangle ECM(SAS)$。$\therefore DM = EM$。$\therefore \triangle DEM$ 是等腰三角形。

7. 如图,在$△ABC$中,$∠C= ∠B,AD⊥BC$,垂足为D,$DE// AB$交AC于点E,若$DE+DC= 4.5$,则$AC+BC$的值为(C)

A. 7.5
B. 8
C. 9
D. 9.5

答案： C