《2025年思维新观察八年级数学下册人教版天津专版》

1.如图,在△ABC中,AB = 5，AC = 13，AD⊥AB交BC于点D，D为BC中点，则AD的长为____.

答案： 延长$AD$至点$E$，使得$AD = ED$，连接$CE$，易证$\triangle ADB\cong\triangle EDC(SAS)$，$AE=\sqrt{13^{2}-5^{2}} = 12$，$\therefore AD = 6$.

2.如图,M是Rt△ABC斜边AB的中点，过点M作DM⊥CM，交AC于点D.若AD = 3，BC = 4，求CD的长.

答案： 延长$DM$至点$E$，使$ME = MD$，连接$BE$，易证$\triangle ADM\cong\triangle BEM$，$\therefore BE = AD = 3$，$BE// AD$，$\therefore CD = CE=\sqrt{3^{2}+4^{2}} = 5$.

3.如图,△ABC中,∠A = 60°，D是BC的中点，E、F分别在AB、AC上，且DE⊥DF，BE = 2，CF = 4，则EF的长为______.

答案： 延长$ED$至点$H$，使得$ED = DH$，连接$FH$，$CH$，过点$F$作$FG\perp HC$于点$G$，易证$\triangle BDE\cong\triangle CDH(SAS)$，$FE = FH$，$CH = BE = 2$，$\angle HCA = 120^{\circ}$，$\therefore\angle FCG = 60^{\circ}$，$\therefore CG = 2$，$FG = 2\sqrt{3}$，$\therefore EF = FH=\sqrt{HG^{2}+FG^{2}} = 2\sqrt{7}$.

4.如图,在△ABC中,∠ABC = 90°，D为BC的中点，点E在AB上，AD、CE交于点F，且AE = EF.若AB = 9，AC = 15，则AE的长为____.

答案： 延长$AD$到点$M$，使$DM = DF$，连接$BM$.$\because BD = DC$，$\angle BDM=\angle CDF$，$\therefore\triangle BDM\cong\triangle CDF$，$\therefore CF = BM$，$\angle M=\angle CFD$，$\therefore CE// BM$，$\therefore\angle AFE=\angle M$，$\because EA = EF$，$\therefore\angle EAF=\angle EFA$，$\therefore\angle BAM=\angle M$，$\therefore AB = BM = CF = 9$，$\because\angle ABC = 90^{\circ}$，$\therefore BC=\sqrt{AC^{2}-AB^{2}} = 12$.
设$AE = x$，则$EF = x$，$CE = 9 + x$，$BE = 9 - x$，$\therefore(9 + x)^{2}-(9 - x)^{2}=12^{2}$，$\therefore x = 4$，即$AE = 4$.