《2025年自主学习指导课程与测试八年级数学上册人教版》

21. （10分）如图，在一个支架的横杆点O处用一根绳悬挂一个小球A，小球A可以摆动，如图，OA表示小球静止时的位置. 当小球从OA摆到OB位置时，过点B作BD⊥OA于点D. 当小球摆到OC位置时，OB与OC恰好垂直，过点C作CE⊥OA于点E，测得CE=24 cm，OA=OB=OC=30 cm.
（1）试说明OE=BD；
（2）求AD的长.

答案： 21.解:
(1)$\because OB\perp OC$,
$\therefore \angle BOD + \angle COE = 90^{\circ}$.
又$\because CE\perp OA$,$BD\perp OA$,
$\therefore \angle CEO = \angle ODB = 90^{\circ}$.
$\therefore \angle BOD + \angle B = 90^{\circ}$.
$\therefore \angle COE = \angle B$.
在$\triangle COE$和$\triangle OBD$中,
$\begin{cases} \angle CEO = \angle BDO \\ \angle COE = \angle B \\ OC = OB \end{cases}$
$\therefore \triangle COE\cong \triangle OBD(AAS)$.
$\therefore OE = BD$.
(2)$\because \triangle COE\cong \triangle OBD$,
$\therefore CE = OD = 24$ $cm$.
$\because OA = 30$ $cm$,
$\therefore AD = OA - OD = 30 - 24 = 6$ $cm$.

22. (12分)如图，在四边形ABCD中，∠ABC=90°，过点B作BE⊥CD，垂足为点E，过点A作AF⊥BE，垂足为点F，且BE=AF.
(1)∠ADE+∠DAF=180°.
(2)求证：AB=BC.
(3)连接BD，且BD平分∠ABE交AF于点G，探究△BCD的形状并说明理由.

答案： 22.
(1)解:$\because AF\perp BE$,$BE\perp CD$,
$\therefore AF// CD$,
$\therefore \angle ADE + \angle DAF = 180^{\circ}$.
故答案为:$180$.
(2)证明:$\because BE\perp CD$,$AF\perp BE$,
$\therefore \angle AFB = \angle BEC = 90^{\circ}$,
$\therefore \angle ABF + \angle BAF = 90^{\circ}$,
$\because \angle ABC = 90^{\circ}$,即$\angle ABF + \angle EBC = 90^{\circ}$,
$\therefore \angle BAF = \angle EBC$,
在$\triangle ABF$和$\triangle BCE$中,$\begin{cases} \angle AFB = \angle BEC \\ AF = BE \\ \angle BAF = \angle EBC \end{cases}$
$\therefore \triangle ABF\cong \triangle BCE(ASA)$,
$\therefore AB = BC$.
(3)解:$\triangle BCD$是等腰三角形,理由如下:
$\because BD$平分$\angle ABE$,
$\therefore \angle ABG = \angle DBE$,
由
(2)知:$\triangle ABF\cong \triangle BCE$,
$\therefore \angle BAF = \angle EBC$,
$\because \angle BGF = \angle BAF + \angle ABG$,$\angle DBC = \angle EBC + \angle DBE$,
$\therefore \angle BGF = \angle DBC$,
$\because BE\perp CD$,$AF\perp BE$,
$\therefore AF// CD$,
$\therefore \angle BGF = \angle BDC$.
$\therefore \angle BDC = \angle DBC$,
$\therefore BC = CD$,
$\therefore \triangle BCD$是等腰三角形.