《2025年一线调研学业测评八年级数学下册北师大版》

1. 如图，在△ABC中，AB = AC = 5，BC = 6，M为BC的中点，MN⊥AC于点N，求MN的长.

答案： 解：连接$AM$，$\because AB = AC = 5$，$BC = 6$，$M$为$BC$的中点，
$\therefore CM=\frac{1}{2}BC = 3$，$AM\perp BC$.$\therefore AM=\sqrt{AC^{2}-CM^{2}}=\sqrt{5^{2}-3^{2}} = 4$.$\because S_{\triangle AMC}=\frac{1}{2}AM\cdot MC=\frac{1}{2}AC\cdot MN$，
$\therefore\frac{1}{2}\times4\times3=\frac{1}{2}\times5MN$，解得$MN = 2.4$.

2. 如图，在△ABC中，AB = AC，D是BC的中点，过点A作EF//BC，且AE = AF，求证：DE = DF.

答案： 证明：连接$AD$.$\because AB = AC$，$D$是$BC$的中点，$\therefore AD\perp BC$. 又$\because EF// BC$，$\therefore AD\perp EF$.$\therefore\angle DAE=\angle DAF = 90^{\circ}$. 又$\because AE = AF$，$AD = AD$，$\therefore\triangle ADE\cong\triangle ADF(SAS)$.$\therefore DE = DF$.

3. 如图，在△ABC中，AC = 2AB，AD平分∠BAC，E是AD上的一点，且EA = EC. 求证：EB⊥AB.

答案：
证明：如图，过点$E$作$EF\perp AC$于点$F$，则$\angle AFE = 90^{\circ}$.

$\because EA = EC$，$\therefore AF = FC=\frac{1}{2}AC$. 又$\because AC = 2AB$，
$\therefore AB = AF$.$\because AD$平分$\angle BAC$，$\therefore\angle BAE=\angle FAE$. 又
$\because AE = AE$，$\therefore\triangle ABE\cong\triangle AFE(SAS)$.$\therefore\angle ABE = \angle AFE = 90^{\circ}$，即$EB\perp AB$.

4. 在四边形ABCD中，AC = BC = BD，AC⊥BD，若AB = √5，求△ABD的面积.

答案：
解：过点$D$作$DE\perp AB$交$BA$的延长
线于点$E$，过点$C$作$CF\perp AB$交$AB$
于点$F$.$\because AC\perp BD$，$CF\perp AB$，
$\therefore\angle ACF+\angle FAC = 90^{\circ}$，$\angle ABD+\angle BAC = 90^{\circ}$.$\therefore\angle ACF = \angle ABD$.
$\because AC = BC$，$CF\perp AB$，$\therefore AF = BF=\frac{\sqrt{5}}{2}$，$\angle ACF = \angle BCF$.$\therefore\angle ABD = \angle BCF$.$\because\angle DEB=\angle AFC = 90^{\circ}$，$\angle ABD=\angle BCF$，$BC = BD$，$\therefore\triangle BDE\cong\triangle CBF(AAS)$.
$\therefore BF = ED=\frac{\sqrt{5}}{2}$.$\therefore\triangle ABD$的面积$=\frac{1}{2}AB\cdot DE=\frac{1}{2}\times\sqrt{5}\times\frac{\sqrt{5}}{2}=\frac{5}{4}$.

5. 转化思想 如图，在等腰直角三角形ABC中，AB = AC，∠BAC = 90°，BF平分∠ABC，CD⊥BF交BF的延长线于点D. 试说明BF = 2CD.

答案：
解：如图，延长$BA$，$CD$相交于点$E$，
则有$\angle BDC=\angle BDE = 90^{\circ}=\angle BAC$.$\because BF$平分$\angle ABC$，
$\therefore\angle CBD=\angle EBD$.$\therefore\angle BCD = \angle E$.$\therefore BC = BE$. 又$\because CE\perp BD$，
$\therefore CD = ED$.$\therefore CE = 2CD$.$\because\angle BAC=\angle BDC = 90^{\circ}$，$\angle AFB=\angle DFC$，$\therefore\angle ABF=\angle ACE$. 又$\because AB = AC$，$\angle BAF=\angle CAE$，$\therefore\triangle ABF\cong\triangle ACE(ASA)$.$\therefore BF = CE$.$\therefore BF = 2CD$.