《2025年经纶学典课时作业八年级数学上册苏科版》

1. 如图,在$\triangle ABC$中,$E是AC$的中点,点$D在BC$上,且$BD = 2CD$,$AD与BE交于点F$,若$S_{\triangle ABF}-S_{四边形CDFE}= \frac{5}{2}$,求$\triangle ABC$的面积.

答案： 设△ABC的面积为a,
∵E是AC的中点,
∴S△ABE = $\frac{1}{2}S_{\triangle ABC}=\frac{1}{2}a$.
∵BD = 2CD,
∴BC = 3CD,
∴S△ACD = $\frac{1}{3}S_{\triangle ABC}=\frac{1}{3}a$,
∴S△ABE - S△ACD = S△ABF - S四边形CDFE = $\frac{1}{2}a - \frac{1}{3}a=\frac{1}{6}a$.
∵S△ABF - S四边形CDFE = $\frac{5}{2}$,
∴$\frac{1}{6}a=\frac{5}{2}$,解得a = 15,即△ABC的面积为15.

2. 如图,在$\triangle ABC$中,点$E是AC$上一点,连接$BE$,且$AB\perp BE$,$CE = 2AE$,过点$C作CD\perp BE$,交$BE的延长线于D$点,已知$BE = 1$,$S_{\triangle CDE}= 4$.
(1)求$DE$的长度;
(2)求$\triangle ABC$的面积.

答案：
(1)连接AD,
∵CE = 2AE,
∴$\frac{S_{\triangle EBC}}{S_{\triangle ABE}}=2$.
∵AB⊥BE,CD⊥BE,
∴AB//CD,
∴S△ABD = S△ABC,
∴S△AED = S△BCE,
∴$\frac{S_{\triangle AED}}{S_{\triangle ABE}}=\frac{S_{\triangle EBC}}{S_{\triangle ABE}}=2$,
∴$\frac{DE}{BE}=\frac{S_{\triangle AED}}{S_{\triangle ABE}} = 2$,
∴DE = 2BE = 2.
(2)由
(1)可知,$\frac{DE}{BE}=2$,即$\frac{S_{\triangle CDE}}{S_{\triangle CBE}}=2$.
∵S△CDE = 4,
∴S△CBE = 2.
∵$\frac{S_{\triangle CBE}}{S_{\triangle ABE}}=2$,
∴S△ABE = 1,
∴S△ABC = S△ABE + S△CBE = 3.

3. 如图,在$\triangle ABC$中,$AD$是它的角平分线.
求证:$AB:AC = BD:CD$.

答案： 作DE⊥AB,DF⊥AC,AH⊥BC,垂足分别为E,F,H.
∵AD平分∠BAC,
∴DE = DF.
∵S△ABD:S△ACD = ($\frac{1}{2}AB\cdot DE$):($\frac{1}{2}AC\cdot DF$),S△ABD:S△ACD = ($\frac{1}{2}BD\cdot AH$):($\frac{1}{2}CD\cdot AH$),
∴AB:AC = BD:CD.

4. 如图,在$\triangle ABC$中,$E为AC$的中点,$AD平分\angle BAC$,$AB:AC = 3:4$,$AD与BE相交于点O$,若$\triangle OAE的面积比\triangle BOD的面积大1$,求$\triangle ADC$的面积.

答案： 如图,作DM⊥AC于M,DN⊥AB于N.
∵AD平分∠BAC,
∴DM = DN,
∴S△ABD:S△ADC = ($\frac{1}{2}AB\cdot DN$):($\frac{1}{2}AC\cdot DM$)=AB:AC = 3:4.设△ABC的面积为a,则S△ADC = $\frac{4}{7}a$.
∵E为AC的中点,
∴S△BEC = $\frac{1}{2}S_{\triangle ABC}=\frac{1}{2}a$.
∵△OAE的面积比△BOD的面积大1,
∴△ADC的面积比△BEC的面积大1,
∴$\frac{4}{7}a - \frac{1}{2}a = 1$,
∴a = 14,
∴S△ADC = $\frac{4}{7}×14 = 8$.