《2025年课堂点睛八年级数学上册人教版》

1. 如图,在四边形 $ ABCD $ 中,$ AB = CD $,$ AD = BC $. 求证:$ ∠A = ∠C $.

答案： 证明:连接 BD.在△ABD 和△CDB中,$\left\{\begin{array}{l} AB=CD,\\ AD=CB,\\ BD=DB,\end{array}\right. \therefore △ABD\cong △CDB$(SSS),$\therefore ∠A=∠C.$

2. 如图,在四边形 $ ABCD $ 中,$ AB = AD $,$ CB = CD $. 求证:$ ∠B = ∠D $.

答案： 证明:连接 AC.在△ABC 和△ADC中,$\left\{\begin{array}{l} AB=AD,\\ CB=CD,\\ AC=AC,\end{array}\right. \therefore △ABC\cong △ADC$(SSS),$\therefore ∠B=∠D.$

3. 如图,线段 $ AC $,$ BD $ 相交于点 $ E $,$ AB = DC $,$ AC = DB $. 求证:$ AE = DE $.

答案： 证明:连接 BC.在△ABC 和△DCB中,$\left\{\begin{array}{l} AB=DC,\\ AC=DB,\\ BC=CB,\end{array}\right. \therefore △ABC\cong △DCB$(SSS),$\therefore ∠A=∠D$.在△AEB 和$△DEC$中,$\left\{\begin{array}{l} ∠A=∠D,\\ ∠AEB=∠DEC,\\ AB=DC,\end{array}\right. \therefore$$△AEB\cong △DEC(AAS),\therefore AE=DE.$

4. 如图,$ ∠ACB = ∠ADE = 90^{\circ} $,$ AC = AD $,$ ∠BAC = ∠EAD $,$ BC $ 的延长线交 $ DE $ 于点 $ F $. 求证:$ BC = CF + EF $.

答案： 证明:连接 AF.$\because ∠ACB=∠ADE$$=90^{\circ },AC=AD,∠BAC=∠EAD,$$\therefore △ABC\cong △AED(ASA),\therefore BC=DE$.在$Rt△ACF$与$Rt△ADF$中,$\left\{\begin{array}{l} AF=AF,\\ AC=AD,\end{array}\right. \therefore Rt△ACF\cong △ADF$(HL),$\therefore CF=DF,\therefore BC=DE=DF$$+EF=CF+EF.$