《2025年全优课堂八年级数学下册》

20. 如图,在□ABCD中,线段EF分别交AD,AC,BC于点E,O,F,EF⊥AC,AO=CO.
(1)求证:△ABF≌△CDE;
(2)在本题的已知条件中,有一个条件如果去掉,并不影响(1)的证明,你认为这个多余的条件是________(直接写出这个条件).

答案： 解:
(1)证明:
∵ 四边形$ABCD$是平行四边形,$\therefore AB = CD,∠B = ∠D,AD = BC,AD// BC$,$\therefore∠EAO = ∠FCO$,在$\triangle AOE$和$\triangle COF$中,$\begin{cases}∠EAO = ∠FCO,\\AO = CO,\\∠AOE = ∠COF,\end{cases}$$\therefore\triangle AOE\cong\triangle COF(ASA)$,$\therefore AE = CF$,$\therefore AD - AE = BC - CF$,即$DE = BF$.在$\triangle ABF$和$\triangle CDE$中,$\begin{cases}AB = CD,\\∠B = ∠D,\\BF = DE,\end{cases}$$\therefore\triangle ABF\cong\triangle CDE(SAS)$;
(2)$EF⊥AC$.

21. 如图,O为□ABCD的对角线AC的中点,过点O作一条直线分别与AD,CB交于点M,N,点E,F在直线MN上,且OE=OF.
求证:∠MAE=∠NCF.

答案： 证明:
∵$O$为$AC$的中点,$\therefore OA = OC$,又$\because OE = OF,∠1 = ∠2$,$\therefore\triangle AOE\cong\triangle COF(SAS)$,$\therefore∠EAO = ∠FCO$,$\because$四边形$ABCD$为平行四边形,$\therefore AD// CB$,$\therefore∠DAC = ∠BCA$,$\therefore∠EAO - ∠DAC = ∠FCO - ∠BCA$,即$∠MAE = ∠NCF$.

22. 如图,在□ABCD中,点E是BC边的中点,连接AE并延长与DC的延长线交于点F.
(1)求证:CF=CD;
(2)若AF平分∠BAD,连接DE,试判断DE与AF的位置关系,并说明理由.

答案： 解:
(1)证明:
∵ 四边形$ABCD$是平行四边形,$\therefore AB = CD,AB// CD$,$\because$点$F$为$DC$的延长线上的一点,$\therefore AB// DF$,$\therefore∠BAE = ∠CFE,∠ECF = ∠EBA$,$\because E$为$BC$的中点,$\therefore BE = CE$,在$\triangle BAE$和$\triangle CFE$中,$\begin{cases}∠BAE = ∠CFE,\\∠ABE = ∠FCE,\\BE = CE,\end{cases}$$\therefore\triangle BAE\cong\triangle CFE(AAS)$,$\therefore AB = CF$,$\therefore CF = CD$;
(2)$DE⊥AF$.
理由:
∵$AF$平分$∠BAD$,$\therefore∠BAF = ∠DAF$,$\because∠BAF = ∠AFD$,$\therefore∠DAF = ∠AFD$,$\therefore DA = DF$,又由
(1),知$\triangle BAE\cong\triangle CFE$,$\therefore AE = EF$,$\therefore DE⊥AF$.

23. (湘西州中考)如图所示,在□ABCD中,AC,BD相交于点O,则下列结论中错误的是 ( )


A. OA=OC
B. ∠ABC=∠ADC
C. AB=CD
D. AC=BD

答案： D

24. (柳州中考)如图,在□ABCD中,全等三角形的对数共有 ( )
A. 2对
B. 3对
C. 4对
D. 5对

答案： C