《2025年轻松作业本八年级数学上册苏科版》

8. 如图,在$△ABC$中,$∠B= 50^{\circ },CD⊥AB$于点D,$∠BCD和∠BDC$的角平分线相交于点E,F为边AC的中点,$CD= CF$,则$∠ACD+∠CED$等于(C)

A. $125^{\circ }$
B. $145^{\circ }$
C. $175^{\circ }$
D. $190^{\circ }$

答案： C

9. 如图,AD是$△ABC$的角平分线,$DE⊥AC$,垂足为E,$BF// AC$交ED的延长线于点F,若BC恰好平分$∠ABF,AE= 2BF$．有下列四个结论：①$DE= DF$；②$DB= DC$；③$AD⊥BC$；④$AB= 3BF$．其中正确的结论有(A)

A. 4个
B. 3个
C. 2个
D. 1个

答案： A

10. 如图,在$Rt△ABC$中,$∠ACB= 90^{\circ }$,D为AB的中点,$∠B= 30^{\circ }$,点E在BC上,且$CE= AC$,则$∠CDE$的大小为______75$^{\circ }$．

答案： 75

11. 如图,长方形ABCD中,$AB= 9,AD= 4$．E为CD边上一点,$CE= 6$．点P从点B出发,以每秒1个单位长度的速度沿着边BA向终点A运动,连接PE．设点P运动的时间为t秒．当$t= $3 或 4 或 $\frac{29}{6}$秒时,$△PAE$是等腰三角形．

答案： 3 或 4 或 $\frac{29}{6}$

12. 如图,在$△ABC$中,AD是边BC上的高,CE是边AB上的中线,G是CE的中点,$DG⊥CE$于点G．求证：$∠B= 2∠BCE$．
证明：连接DE。$\because G$ 是 $CE$ 的中点，$DG \perp CE$，$\therefore DG$ 是 $CE$ 的垂直平分线。$\therefore$DE=DC。$\therefore$∠DEC=∠DCE。$\because$∠BDE=∠DEC+∠DCE，$\therefore \angle BDE = 2\angle BCE$。$\because CE$ 是边 $AB$ 上的中线，$\therefore E$ 是 $BA$ 中点，即AE=BE。在 $Rt\triangle ABD$ 中，DE=AE=BE。$\therefore$∠BDE=∠B。$\therefore \angle B = 2\angle BCE$。

答案： 证明：连接 $DE$。$\because G$ 是 $CE$ 的中点，$DG \perp CE$，$\therefore DG$ 是 $CE$ 的垂直平分线。$\therefore DE = DC$。$\therefore \angle DEC = \angle DCE$。$\because \angle BDE = \angle DEC + \angle DCE$，$\therefore \angle BDE = 2\angle BCE$。$\because CE$ 是边 $AB$ 上的中线，$\therefore E$ 是 $BA$ 中点，即 $AE = BE$。在 $Rt\triangle ABD$ 中，$DE = AE = BE$。$\therefore \angle BDE = \angle B$。$\therefore \angle B = 2\angle BCE$。

13. 如图,在$△ABC$中,CE、BD分别是AB、AC边上的高线,M是BC的中点,连接DE、EM、MD．
(1)求证：$ME= MD$；
(2)若$∠A= 45^{\circ }$,求$∠EDM$的度数．

(1)证明：$\because CE$、$BD$ 分别是 $AB$、$AC$ 边上的高线，$\therefore \angle BEC = \angle CDB = 90^{\circ}$。$\because M$ 是 $BC$ 的中点，$\therefore EM = \frac{1}{2}BC$，$DM = \frac{1}{2}BC$。$\therefore ME = MD$；
(2)解：$\because \angle A = 45^{\circ}$，$\therefore \angle ABC + \angle ACB = 135^{\circ}$。$\because EM = BM$，$DM = CM$，$\therefore \angle BEM = \angle ABC$，$\angle MDC = \angle ACB$。$\therefore \angle EBM + \angle BEM + \angle ACB + \angle MDC = 135^{\circ} × 2 = 270^{\circ}$。$\therefore \angle EMB + \angle DMC = 180^{\circ} × 2 - 270^{\circ} = 90^{\circ}$。$\therefore \angle EMD = 90^{\circ}$。$\because ME = MD$，$\therefore \angle EDM =$$45^{\circ}$。

答案： （1）证明：$\because CE$、$BD$ 分别是 $AB$、$AC$ 边上的高线，$\therefore \angle BEC = \angle CDB = 90^{\circ}$。$\because M$ 是 $BC$ 的中点，$\therefore EM = \frac{1}{2}BC$，$DM = \frac{1}{2}BC$。$\therefore ME = MD$；
（2）解：$\because \angle A = 45^{\circ}$，$\therefore \angle ABC + \angle ACB = 135^{\circ}$。$\because EM = BM$，$DM = CM$，$\therefore \angle BEM = \angle ABC$，$\angle MDC = \angle ACB$。$\therefore \angle EBM + \angle BEM + \angle ACB + \angle MDC = 135^{\circ} \times 2 = 270^{\circ}$。$\therefore \angle EMB + \angle DMC = 180^{\circ} \times 2 - 270^{\circ} = 90^{\circ}$。$\therefore \angle EMD = 90^{\circ}$。$\because ME = MD$，$\therefore \angle EDM = 45^{\circ}$。