《2025年1加1轻巧夺冠完美期末八年级数学上册人教版辽宁专版》

18. (8分)(抚顺清原期末)
如图,在$\triangle A B C$中,$A D$平分$\angle B A C$交$B C$于点$D,B E$平分$\angle A B C$交$A D$于点$E$.
(1)若$\angle C=80^{\circ},\angle B A C=60^{\circ}$,求$\angle A D B$的度数.
(2)若$\angle B E D=65^{\circ}$,求$\angle C$的度数.

答案： 18.解：
(1)$\because AD$平分$\angle BAC$，$\angle BAC = 60^{\circ}$，$\therefore \angle DAC = \frac{1}{2} \angle BAC = 30^{\circ}$，$\because \angle ADB$是$\triangle ADC$的一个外角，$\angle C = 80^{\circ}$，$\therefore \angle ADB = \angle C + \angle DAC = 80^{\circ} + 30^{\circ} = 110^{\circ}$.
(2)$\because AD$平分$\angle BAC$，$BE$平分$\angle ABC$，$\therefore \angle BAC = 2\angle BAD$，$\angle ABC = 2\angle ABE$，$\because \angle BED$是$\triangle ABE$的一个外角，$\angle BED = 65^{\circ}$，$\therefore \angle BAD + \angle ABE = \angle BED = 65^{\circ}$，$\therefore \angle BAC + \angle ABC = 2(\angle BAD + \angle ABE) = 130^{\circ}$，$\because \angle BAC + \angle ABC + \angle C = 180^{\circ}$，$\therefore \angle C = 180^{\circ} - (\angle BAC + \angle ABC) = 50^{\circ}$.

19. (8分)
(1)如图1,在$\triangle A B C$中,点$M$在$C B$的延长线上,点$N$在线段$A C$上,连接$M N$交$A B$于点$D,\angle B A N$和$\angle C M N$的平分线交于点$P$.猜想$\angle C,\angle B D N$和$\angle P$之间的数量关系并证明.
(2)如图2,在$\triangle A B C$中,点$M$在线段$C B$上,点$N$在$C A$的延长线上,连接$M N$交$A B$于点$D,\angle B A N$和$\angle B M N$的平分线交于点$P$,探究$\angle C,\angle B D N$和$\angle P$之间的数量关系,并说明理由.

答案：
19.解：
(1)$\angle C + \angle BDN = 2\angle P$，理由如下：延长$AP$交$BC$于点$E$，如答图1，$\because \angle BAN$和$\angle CMN$的平分线交于点$P$，$\therefore$设$\angle PAB = \angle PAC = \alpha$，$\angle PMN = \angle PMC = \beta$，则$\angle BAC = 2\alpha$，$\angle NMC = 2\beta$，$\because \angle MNA$是$\triangle MNC$的一个外角，$\therefore \angle MNA = \angle C + \angle NMC = \angle C + 2\beta$，$\because \angle BDN$是$\triangle AND$的一个外角，$\therefore \angle BDN = \angle BAC + \angle MNA = 2\alpha + \angle C + 2\beta = \angle C + 2(\alpha + \beta)$，$\because \angle PEM$是$\triangle AEC$的一个外角，$\therefore \angle PEM = \angle C + \angle PAC = \angle C + \alpha$，$\because \angle APM$是$\triangle PME$的一个外角，$\therefore \angle APM = \angle PEM + \angle PMC = \angle C + \alpha + \beta$，$\therefore \alpha + \beta = \angle APM - \angle C$，$\therefore \angle BDN = \angle C + 2(\alpha + \beta) = \angle C + 2(\angle APM - \angle C)$，$\therefore \angle C + \angle BDN = 2\angle APM$.
　　　　　　　　 　　
(2)$\angle BDN - \angle C = 2\angle P$，理由如下：设$MN$与$PA$交于点$F$，如答图2，$\because \angle BAN$和$\angle BMN$的平分线交于点$P$，$\therefore$设$\angle PAB = \angle PAN = \alpha$，$\angle AMB = \angle PMN = \beta$，则$\angle BAN = 2\alpha$，$\angle BMN = 2\beta$，$\therefore \angle BAC = 180^{\circ} - \angle BAN = 180^{\circ} - 2\alpha$，$\because \angle PFN$是$\triangle PFM$和$\triangle AFN$的一个外角，$\therefore \angle PFN = \angle P + \angle PMN = \angle N + \angle PAN$，$\therefore \angle P + \beta = \angle N + \alpha$，$\therefore \angle N = \angle P + \beta - \alpha$，$\because \angle BMN$是$\triangle CMN$的一个外角，$\therefore \angle BMN = \angle C + \angle N$，$\therefore 2\beta = \angle C + \angle P + \beta - \alpha$，$\therefore \alpha + \beta = \angle C + \angle P$，$\because \angle BDN$为$\triangle ADN$的一个外角，$\therefore \angle BDN = \angle N + \angle BAN = \angle P + \beta - \alpha + 2\alpha = \angle P + \alpha + \beta = \angle P + \angle C + \angle P$，$\therefore \angle BDN - \angle C = 2\angle P$.