《2025年启东中学作业本八年级数学上册苏科版连淮专版》

1. (2024·连云区期末)如图,$AB= AC$,$BD= CD$,$∠A= 60^{\circ}$,$∠D= 140^{\circ}$,则$∠B= $ (B)

A.$30^{\circ}$
B.$40^{\circ}$
C.$50^{\circ}$
D.$70^{\circ}$

答案： B

2. (2024·清江浦区期中)如图,$∠A= ∠C= 90^{\circ}$,$AB= CD$. 若$OB= 2OA= 6$,则$BC$的长为______9.

答案： 9

3. 如图,$AB= AE$,$∠B= ∠E$,$BC= ED$,$AF平分∠BAE$,求证：$AF⊥CD$.

答案：
证明:如答图，连接AC,AD.
在$\triangle ABC$和$\triangle AED$中,$\left\{\begin{array}{l} AB=AE,\\ ∠B=∠E,\\ BC=ED,\end{array}\right. $
$\therefore \triangle ABC\cong \triangle AED(SAS),\therefore AC=AD,∠BAC=∠EAD.$
$\because AF$平分$∠BAE,\therefore ∠BAF=∠EAF,$
$\therefore ∠BAF - ∠BAC = ∠EAF - ∠EAD,$
即$∠FAC = ∠FAD.$
在$\triangle ACF$和$\triangle ADF$中,$\left\{\begin{array}{l} AC=AD,\\ ∠CAF=∠DAF,\\ AF=AF,\end{array}\right. $
$\therefore \triangle ACF\cong \triangle ADF(SAS),\therefore ∠AFC = ∠AFD.$
$\because ∠AFC + ∠AFD = 180^{\circ },\therefore ∠AFC = 90^{\circ },$
即$AF⊥CD.$

4. 如图,$AB= AE$,$AB⊥AE$,$AD= AC$,$AD⊥AC$,$M为BC$的中点,求证：$DE= 2AM$.

答案：
证明:如答图，延长AM至点N,使$MN = AM$,连接BN.

$\because M$为$BC$的中点,$\therefore CM = BM.$
在$\triangle AMC$和$\triangle NMB$中,$\left\{\begin{array}{l} AM=NM,\\ ∠AMC=∠NMB,\\ CM=BM,\end{array}\right. $
$\therefore \triangle AMC\cong \triangle NMB(SAS),\therefore AC = BN,∠C = ∠NBM.$
$\because AD = AC,\therefore AD = BN.$
$\because AB⊥AE,AD⊥AC,\therefore ∠EAB = ∠DAC = 90^{\circ },$
$\therefore ∠EAD + ∠BAC = 180^{\circ },$
$\therefore ∠ABN = ∠ABC + ∠C = 180^{\circ } - ∠BAC = ∠EAD.$
在$\triangle EAD$和$\triangle ABN$中,$\left\{\begin{array}{l} AE=BA,\\ ∠EAD=∠ABN,\\ AD=BN,\end{array}\right. $
$\therefore \triangle EAD\cong \triangle ABN(SAS),\therefore DE = AN = 2AM.$

5. 如图,$CE$,$CB分别是\triangle ABC与\triangle ADC$的中线,且$∠ACB= ∠ABC$. 求证：$CD= 2CE$.

答案：
证明:如答图，延长CE至点F,使$EF = CE$,连接BF.
$\therefore CF = 2CE.$

$\because CE$是$\triangle ABC$的中线,$\therefore AE = BE.$
在$\triangle ACE$和$\triangle BFE$中,$\left\{\begin{array}{l} CE=FE,\\ ∠AEC=∠BEF,\\ AE=BE,\end{array}\right. $
$\therefore \triangle ACE\cong \triangle BFE(SAS),\therefore ∠A = ∠FBE,AC = BF.$
又$\because ∠ACB = ∠ABC,CB$是$\triangle ADC$的中线,
$\therefore AC = AB = BD = BF.$
$\because ∠DBC = ∠A + ∠ACB = ∠ABF + ∠ABC,$
$\therefore ∠DBC = ∠FBC.$
在$\triangle DBC$和$\triangle FBC$中,$\left\{\begin{array}{l} DB=FB,\\ ∠DBC=∠FBC,\\ BC=BC,\end{array}\right. $
$\therefore \triangle DBC\cong \triangle FBC(SAS),\therefore DC = CF = 2CE.$