《2025年一线调研学业测评八年级数学下册北师大版》

11. (苏州中考)定义：一个三角形的一边长是另一边长的2倍，这样的三角形叫做“倍长三角形”. 若等腰△ABC是“倍长三角形”，底边BC的长为3，则腰AB的长为_____.

答案： 6

12. (绍兴中考)如图，在△ABC中，AB = AC，∠B = 70°，以点C为圆心、CA长为半径作弧，交直线BC于点P，连接AP，则∠BAP的度数是___________.

答案： 15°或 75°

13. (深圳中考)如图，已知∠BAC = 60°，AD是角平分线，且AD = 10，作AD的垂直平分线交AC于点F，作DE⊥AC，则△DEF的周长为__________.

答案： $5 + 5\sqrt{3}$

14. (长沙中考)如图，在△ABC中，AD⊥BC，垂足为点D，BD = CD，延长BC至点E，使得CE = CA，连接AE.
(1)求证：∠B = ∠ACB.
(2)若AB = 5，AD = 4，求△ABE的周长和面积.

答案：
(1)证明：$\because AD\perp BC,BD = CD$，
$\therefore AD$是线段$BC$的垂直平分线.
$\therefore AB = AC.\therefore\angle B=\angle ACB$.
(2)解：由
(1)得$CE = AC = AB = 5$.
在$Rt\triangle ADB$中，$BD=\sqrt{AB^{2}-AD^{2}}=\sqrt{5^{2}-4^{2}} = 3$，
$\therefore CD = BD = 3.\therefore BE = 2BD + CE = 2\times3 + 5 = 11$. 在
$Rt\triangle ADE$中，$AE=\sqrt{AD^{2}+DE^{2}}=\sqrt{4^{2}+8^{2}} = 4\sqrt{5}$.
$\therefore C_{\triangle ABE}=AB + BE + AE = 16 + 4\sqrt{5},S_{\triangle ABE}=\frac{1}{2}BE\cdot AD=\frac{1}{2}\times11\times4 = 22$.

15. 如图，在△ABC中，AB = AC，AD⊥BC于点D.
(1)若∠C = 42°，求∠BAD的度数.
(2)若点E在边AB上，EF//AC交AD的延长线于点F. 求证：AE = FE.

答案：
(1)解：$\because AB = AC,AD\perp BC,\therefore\angle BAD=\angle CAD$，$\angle ADC = 90^{\circ}$. 又$\because\angle C = 42^{\circ}$，$\therefore\angle BAD=\angle CAD = 90^{\circ}-42^{\circ}=48^{\circ}$.
(2)证明：$\because AB = AC,AD\perp BC,\therefore\angle BAD=\angle CAD$.
$\because EF// AC,\therefore\angle F=\angle CAD.\therefore\angle BAD=\angle F$.
$\therefore AE = FE$.

16. 如图，AD是△ABC的高，AD的垂直平分线分别交AB，AD，AC于点E，H，F.
(1)求证：∠B = $\frac{1}{2}$∠AED.
(2)若DE = 1，求AB的长.

答案：
(1)证明：$\because EF$是$AD$的垂直平分线，$\therefore EA = ED$.
$\because EH\perp AD,\therefore\angle AEH=\angle DEH.\because EF\perp AD,BC\perp AD.\therefore EF// BC.\therefore\angle AEH=\angle B.\therefore\angle B=\frac{1}{2}\angle AED$.
(2)解：由
(1)得$EF// BC,\therefore\angle HED=\angle EDB$.
$\because\angle AEH=\angle HED,\angle AEH=\angle B,\therefore\angle B=\angle EDB$.
$\therefore BE = DE = AE.\therefore AB = 2BE = 2DE = 2\times1 = 2$.

17. (杭州中考)如图，在△ABC中，∠ABC的平分线BD交AC于点D，AE⊥BC于点E. 已知∠ABC = 60°，∠C = 45°.
(1)求证：AB = BD.
(2)若AE = 3，则△ABC的面积为________.

答案：
(1)证明：$\because BD$平分$\angle ABC,\angle ABC = 60^{\circ},\therefore\angle DBC=\frac{1}{2}\angle ABC = 30^{\circ}$. $\because\angle C = 45^{\circ},\therefore\angle ADB=\angle DBC+\angle C = 75^{\circ}$. $\angle BAC = 180^{\circ}-\angle ABC-\angle C = 75^{\circ}$.
$\therefore\angle BAC=\angle ADB.\therefore AB = BD$.
(2)$\frac{9 + 3\sqrt{3}}{2}$ 提示：在$Rt\triangle ABE$中，$\angle ABC = 60^{\circ}$，
$\therefore\angle BAE = 30^{\circ}.\therefore AB = 2BE$. 又$\because AE = 3,\therefore BE=\sqrt{3}$.
在$Rt\triangle AEC$中，$\angle C = 45^{\circ},AE = 3,\therefore EC = 3.\therefore BC = 3+\sqrt{3}.\therefore S_{\triangle ABC}=\frac{1}{2}BC\cdot AE=\frac{9 + 3\sqrt{3}}{2}$.