《2025年自主学习指导课程与测试八年级数学上册人教版》

23. （12分）如图甲，在△ABC中，AE⊥BC于E，AE=BE，D是AE上的一点，且DE=CE，连接BD，CD.
（1）试判断BD与AC的位置关系和数量关系，并说明理由.
（2）如图乙，若将△DCE绕点E旋转一定的角度后，试判断BD与AC的位置关系和数量关系是否发生变化，并说明理由.
（3）如图丙，若将图乙中的等腰直角三角形都换成等边三角形，其他条件不变.
①试猜想BD与AC的数量关系，并说明理由；
②你能求出BD与AC所夹的锐角度数吗？如果能，请直接写出这个角的度数；如果不能，请说明理由.

答案：
23.解:
(1)$BD = AC$,$BD\perp AC$,

理由:延长$BD$交$AC$于$F$.
$\because AE\perp BC$,$\therefore \angle AEB = \angle AEC = 90^{\circ}$.
在$\triangle BED$和$\triangle AEC$中$\begin{cases} BE = AE \\ \angle BED = \angle AEC \\ DE = EC \end{cases}$
$\therefore \triangle BED\cong \triangle AEC$.
$\therefore BD = AC$,$\angle DBE = \angle CAE$.
$\because \angle BED = 90^{\circ}$,
$\therefore \angle EBD + \angle BDE = 90^{\circ}$.
$\because \angle BDE = \angle ADF$,
$\therefore \angle ADF + \angle CAE = 90^{\circ}$.
$\therefore \angle AFD = 180^{\circ}-90^{\circ}=90^{\circ}$.
$\therefore BD\perp AC$.
(2)不发生变化.
理由:$\because \angle BEA = \angle DEC = 90^{\circ}$,
$\therefore \angle BEA + \angle AED = \angle DEC + \angle AED$.

$\therefore \angle BED = \angle AEC$.
在$\triangle BED$和$\triangle AEC$中,
$\begin{cases} BE = AE \\ \angle BED = \angle AEC \\ DE = EC \end{cases}$
$\therefore \triangle BED\cong \triangle AEC(SAS)$.
$\therefore BD = AC$,$\angle BDE = \angle ACE$.
$\because \angle DEC = 90^{\circ}$,$\therefore \angle ACE + \angle EOC = 90^{\circ}$.
$\because \angle EOC = \angle DOF$,$\therefore \angle BDE + \angle DOF = 90^{\circ}$.
$\therefore \angle DFO = 180^{\circ}-90^{\circ}=90^{\circ}$.
$\therefore BD\perp AC$.
(3)①$\because \angle BEA = \angle DEC = 90^{\circ}$,
$\therefore \angle BEA + \angle AED = \angle DEC + \angle AED$.
$\therefore \angle BED = \angle AEC$.
在$\triangle BED$和$\triangle AEC$中,
$\begin{cases} BE = AE \\ \angle BED = \angle AEC \\ DE = EC \end{cases}$
$\therefore \triangle BED\cong \triangle AEC(SAS)$.
$\therefore BD = AC$.
②能.
理由:$\because \triangle ABE$和$\triangle DEC$是等边三角形,
$\therefore AE = BE$,$DE = EC$,$\angle EDC = \angle DCE = 60^{\circ}$,
$\angle BEA = \angle DEC = 60^{\circ}$.
$\therefore \angle BEA + \angle AED = \angle DEC + \angle AED$.
$\therefore \angle BED = \angle AEC$.
在$\triangle BED$和$\triangle AEC$中,
$\begin{cases} BE = AE \\ \angle BED = \angle AEC \\ DE = EC \end{cases}$
$\therefore \triangle BED\cong \triangle AEC(SAS)$.
$\therefore \angle BDE = \angle ACE$,$BD = AC$.
$\therefore \angle DFC = 180^{\circ}-(\angle BDE + \angle EDC + \angle DCF)$
$=180^{\circ}-(60^{\circ}+60^{\circ})$
$=60^{\circ}$,
即$BD$与$AC$所夹的锐角的度数为$60^{\circ}$.