《2025年一线调研学业测评八年级数学下册北师大版》

8. 如图，在$\triangle ABC$中，点D是BC上一点，且$BD = DA = AC$. 把边AB绕着点A顺时针旋转一定角度得到$\angle BAE$，连接DE，交AB于点F.
(1)若$\angle B = \alpha$，请用含$\alpha$的式子表示$\angle C$.
(2)若$\angle CAD = \angle BAE$，求证：DA平分$\angle CDE$.

答案：
(1)解：$\because AD = BD$，$\angle B=\alpha$，$\therefore\angle BAD=\angle B=\alpha$.
$\therefore\angle ADC=\angle B+\angle BAD = 2\alpha$.
$\because AD = AC$，$\therefore\angle C=\angle ADC = 2\alpha$.
(2)证明：$\because\angle CAD=\angle BAE$，$\therefore\angle CAB=\angle DAE$.
在$\triangle ABC$和$\triangle AED$中，$\begin{cases}AC = AD,\\\angle CAB=\angle DAE,\\AB = AE,\end{cases}$
$\therefore\triangle ABC\cong\triangle AED(SAS)$.$\therefore\angle C=\angle ADE$.
$\because\angle C=\angle ADC$，$\therefore\angle ADE=\angle ADC$.
$\therefore DA$平分$\angle CDE$.

9. 如图，已知$AC\perp BC$，垂足为C，$AC = 4$，$BC = 3\sqrt{3}$，将线段AC绕点A按逆时针方向旋转$60^{\circ}$，得到线段AD，连接DC，DB.
(1)线段$DC =$______.
(2)求线段DB的长度.

答案：
解：
(1)4
(2)如图，过点$D$作$DE\perp BC$于点$E$. 由旋转的性质，得$\angle A = 60^{\circ}$，$AD = AC$.$\therefore\triangle ACD$是等边三角形.
$\therefore\angle ACD = 60^{\circ}$. 又$\because AC\perp BC$，
$\therefore\angle DCE=\angle ACB-\angle ACD = 30^{\circ}$.
$\therefore$在$Rt\triangle CDE$中，$DE=\frac{1}{2}DC = 2$.
$\therefore CE=\sqrt{DC^{2}-DE^{2}}=2\sqrt{3}$.
$\therefore BE = BC - CE = 3\sqrt{3}-2\sqrt{3}=\sqrt{3}$.
$\therefore$在$Rt\triangle BDE$中，$DB=\sqrt{DE^{2}+BE^{2}}=\sqrt{7}$.

10. 已知$\triangle ABC$与$\triangle DEC$是两个大小不同的等腰直角三角形.
(1)如图①所示，连接AE，DB，试判断线段AE和DB的数量和位置关系，并说明理由.
(2)如图②所示，连接DB，将线段DB绕点D顺时针旋转$90^{\circ}$到DF，连接AF，试判断线段DE和AF的数量和位置关系，并说明理由.

答案：
解：
(1)$AE = DB$，$AE\perp DB$. 理由如下：如图①，延长$DB$交$AE$于点$H$.$\because\triangle ABC$与$\triangle DEC$是等腰直角三角形，$\therefore AC = BC$，$\angle ACE=\angle BCD = 90^{\circ}$，$CE = CD$.
$\therefore\triangle ACE\cong\triangle BCD$.$\therefore AE = DB$，$\angle AEC=\angle BDC$.
又$\because\angle EBH=\angle CBD$，$\therefore\angle BHE=\angle BCD = 90^{\circ}$.
$\therefore AE\perp DB$.
(2)$DE = AF$，$DE\perp AF$. 理由如下：如图②，设$DE$与$AF$交于点$N$.$\because\triangle ABC$与$\triangle DEC$是等腰直角三角形，$\therefore AC = BC$，$\angle ACE=\angle BCD = 90^{\circ}$，$\angle E=\angle EDC = 45^{\circ}$，$CE = CD$.$\therefore CE - BC = CD - AC$，即$BE = DA$.
由旋转的性质，得$DB = DF$，$\angle BDF = 90^{\circ}$.$\because\angle EBD=\angle C+\angle BDC = 90^{\circ}+\angle BDC$，$\angle ADF=\angle BDF+\angle BDC = 90^{\circ}+\angle BDC$，$\therefore\angle EBD=\angle ADF$.$\therefore\triangle EBD\cong\triangle ADF$.
$\therefore DE = AF$，$\angle E=\angle FAD$. 又$\because\angle E=\angle EDC = 45^{\circ}$，
$\therefore\angle FAD = 45^{\circ}$.$\therefore\angle AND = 90^{\circ}$，即$DE\perp AF$.
(图①、图②分别对应 10 题中
(1)、
(2)的图形)