《2025年基础训练大象出版社八年级数学上册人教版》

19. (9分)如图,在$Rt\triangle ABC$中,$\angle B = 90^{\circ}$,$AD是\angle BAC$的平分线,$DE\perp AC于点E$,点$F在AB$上,$DC = DF$.
(1)求证:$BF = CE$;
(2)若$AC = 10$,$AF = 2$,求$CE$的长.

答案：
(1) $\because$ AD是$\angle BAC$的平分线，$\angle B = 90^{\circ}$, $DE\perp AC$,$\therefore DB = DE$, $\angle B = \angle DEC = \angle DEA = 90^{\circ}$.在$Rt\triangle BDF$和$Rt\triangle EDC$中，$\left\{\begin{array}{l} DF = DC\\ DB = DE\end{array}\right.$$\therefore Rt\triangle BDF \cong Rt\triangle EDC(HL)$.$\therefore BF = CE$.
(2) 在$Rt\triangle ABD$和$Rt\triangle AED$中，$\left\{\begin{array}{l} AD = AD\\ DB = DE\end{array}\right.$$\therefore Rt\triangle ABD \cong Rt\triangle AED(HL)$.$\therefore AB = AE$.$\because AC = 10$, $AF = 2$,$\therefore AB = BF + 2$, $AE = 10 - CE$;$\therefore BF + 2 = 10 - CE$.$\therefore CE + 2 = 10 - CE$.$\therefore CE = 4$.

20. (9分)数学活动课上,在学习了角平分线的尺规作图后,嘉嘉受此问题启发,利用轴对称性又发现了一种作角平分线的方法.如图,请仔细阅读并完成相应任务.
【作法】
①以点$O$为圆心,适当长为半径作弧,交$OA于点C$,交$OB于点D$;
②再以点$O$为圆心,大于$OC$长为半径作弧,交$OA于点E$,交$OB于点F$;
③连接$CF$,$DE交于点P$;
④作射线$OP$.
则射线$OP即为\angle AOB$的平分线.
【任务】
(1)由尺规作图可直接得到的相等线段有$OC = OD$和$OE = OF$.
(2)由(1)中的条件,可证$\triangle OCF\cong\triangle ODE$,依据是$SAS$(填判定方法).
(3)如果把(2)中的$\triangle OCF\cong\triangle ODE$作为条件,求证:$OP平分\angle AOB$.
$\because \triangle OCF \cong \triangle ODE$,$\therefore \angle OFC = \angle OED$.$\because OC = OD$, $OE = OF$,$\therefore OE - OC = OF - OD$,即$CE = DF$.在$\triangle PCE$和$\triangle PDF$中，$\left\{\begin{array}{l} \angle PEC = \angle PFD\\ \angle CPE = \angle DPF\\ CE = DF\end{array}\right.$$\therefore \triangle PCE \cong \triangle PDF(AAS)$.$\therefore PE = PF$.在$\triangle POE$和$\triangle POF$中，$\left\{\begin{array}{l} PE = PF\\ OE = OF\\ OP = OP\end{array}\right.$$\therefore \triangle POE \cong \triangle POF(SSS)$.$\therefore \angle POE = \angle POF$,即OP平分$\angle AOB$.

答案：
(1) $OE = OF$
(2) $SAS$
(3) $\because \triangle OCF \cong \triangle ODE$,$\therefore \angle OFC = \angle OED$.$\because OC = OD$, $OE = OF$,$\therefore OE - OC = OF - OD$,即$CE = DF$.在$\triangle PCE$和$\triangle PDF$中，$\left\{\begin{array}{l} \angle PEC = \angle PFD\\ \angle CPE = \angle DPF\\ CE = DF\end{array}\right.$$\therefore \triangle PCE \cong \triangle PDF(AAS)$.$\therefore PE = PF$.在$\triangle POE$和$\triangle POF$中，$\left\{\begin{array}{l} PE = PF\\ OE = OF\\ OP = OP\end{array}\right.$$\therefore \triangle POE \cong \triangle POF(SSS)$.$\therefore \angle POE = \angle POF$,即OP平分$\angle AOB$.