《2025年学霸题中题八年级数学上册人教版》

1. (1)如图①,在$\triangle ABC$中,$∠ABC$与$∠ACB$的平分线交于点$P$,$∠A = 64^{\circ}$,则$∠BPC =$$122^{\circ }$;
(2)如图②,$\triangle ABC$的内角$∠ACB$的平分线与外角$∠ABD$的平分线交于点$E$.$∠A$与$∠BEC$的数量关系为$∠BEC=\frac {1}{2}∠A$;
(3)如图③,$\triangle ABC$的外角$∠CBM$,$∠BCN$的平分线交于点$Q$,$∠BQC$与$∠A$的数量关系为$∠BQC=90^{\circ }-\frac {1}{2}∠A$.若$∠A = 64^{\circ}$,$∠CBQ$,$∠BCQ$的平分线交于点$P$,则$∠BPC =$119$^{\circ}$,$∠ECQ$的平分线与$BP$的延长线相交于点$R$,则$∠R =$29$^{\circ}$.

答案：
(1)$122^{\circ }$
(2)$∠BEC=\frac {1}{2}∠A$
(3)$∠BQC=90^{\circ }-\frac {1}{2}∠A$ 119 29 解析：$∠BQC=90^{\circ }-\frac {1}{2}∠A=90^{\circ }-\frac {1}{2}×64^{\circ }=58^{\circ },∠BPC=90^{\circ }+\frac {1}{2}∠BQC=90^{\circ }+\frac {1}{2}×58^{\circ }=119^{\circ },∠R=\frac {1}{2}∠BQC=29^{\circ }$。

2. 改编题 (1)如图①,$∠A = 60^{\circ}$,$BO$,$CO$分别是$∠ABC$,$∠ACB$的三等分线(即$∠OBC = \frac{1}{3}∠ABC$,$∠OCB = \frac{1}{3}∠ACB$),则$∠BOC$的度数为____140°.
(2)如图②,$∠ABO$,$∠ACO$的十二等分线分别相交于点$O_{1}$,$O_{2}$,$\cdots$,$O_{11}$,若$∠BOC = 115^{\circ}$,$∠BO_{1}C = 60^{\circ}$,则$∠A$的度数是多少?55°

答案：
(1)$140^{\circ }$ 解析：$\because ∠A=60^{\circ },\therefore ∠ABC+∠ACB=180^{\circ }-60^{\circ }=120^{\circ }$。$\because BO$，$CO$分别是$∠ABC$，$∠ACB$的三等分线，$\therefore ∠OBC+∠OCB=\frac {1}{3}×120^{\circ }=40^{\circ }$，$\therefore ∠BOC=180^{\circ }-(∠OBC+∠OCB)=140^{\circ }$。
(2)设$∠ABO=x$，$∠ACO=y$。由题意可得$\begin{cases}x + y + ∠A = 115^{\circ }\\\frac{x}{12}+\frac{y}{12}+∠A = 60^{\circ }\end{cases}$，解得$∠A=55^{\circ }$。

3. 改编题 (1)如图①,直线$AP$平分$∠BAD$,$CP$平分$∠BCD$的外角$∠BCE$,猜想$∠P$与$∠B$,$∠D$的数量关系,并说明理由;
$∠P=$$90^{\circ }+\frac {1}{2}(∠B+∠D)$。理由如下：∵直线AP平分$∠BAD$，CP平分$∠BCD$的外角$∠BCE$，$\therefore ∠PAB=∠PAD=\frac {1}{2}∠BAD$，$∠PCB=∠PCE=\frac {1}{2}∠BCE$，$\therefore 2∠PAB+∠B=180^{\circ }-2∠PCB+∠D$，$\therefore 180^{\circ }-2(∠PAB+∠PCB)+∠D=∠B$。$\because ∠P+∠PAD=∠PCD+∠D$，$∠BAD+∠B=∠BCD+∠D$，$\therefore ∠P+∠PAD-∠BAD-∠B=∠PCD-∠BCD$，$\therefore ∠P-∠PAB-∠B=∠PCB$，$\therefore ∠P-∠B=∠PAB+∠PCB$，$\therefore 180^{\circ }-2(∠P-∠B)+∠D=∠B$，即$∠P=90^{\circ }+\frac {1}{2}(∠B+∠D)$。
(2)如图②,直线$AP$平分$∠BAD$的外角$∠FAD$,$CP$平分$∠BCD$的外角$∠BCE$,猜想$∠P$与$∠B$,$∠D$的数量关系,并说明理由;
$∠P=$$180^{\circ }-\frac {1}{2}(∠B+∠D)$。理由如下：连接PB，PD。∵直线AP平分$∠BAD$的外角$∠FAD$，CP平分$∠BCD$的外角$∠BCE$，$\therefore ∠FAP=∠PAO$，$∠PCE=∠PCB$。$\because ∠APB+∠PBA+∠PAB=180^{\circ }$，$∠PCB+∠PBC+∠BPC=180^{\circ }$，$\therefore ∠APC+∠ABC+∠PCB+∠PAB=360^{\circ }$，同理得$∠APC+∠ADC+∠PCD+∠PAD=360^{\circ }$，$\therefore 2∠APC+∠ABC+∠ADC+∠PCB+∠PAB+∠PCD+∠PAD=720^{\circ }$，$\therefore 2∠APC+∠ABC+∠ADC+∠PCE+∠PAB+∠PCD+∠PAF=720^{\circ }$。$\because ∠PCE+∠PCD=180^{\circ }$，$∠PAB+∠PAF=180^{\circ }$，$\therefore 2∠APC+∠ABC+∠ADC=360^{\circ }$，$\therefore ∠APC=180^{\circ }-\frac {1}{2}(∠ABC+∠ADC)$，即$∠P=180^{\circ }-\frac {1}{2}(∠B+∠D)$。

答案：
(1)$∠P=90^{\circ }+\frac {1}{2}(∠B+∠D)$。理由如下：
∵直线AP平分$∠BAD$，CP平分$∠BCD$的外角$∠BCE$，$\therefore ∠PAB=∠PAD=\frac {1}{2}∠BAD$，$∠PCB=∠PCE=\frac {1}{2}∠BCE$，$\therefore 2∠PAB+∠B=180^{\circ }-2∠PCB+∠D$，$\therefore 180^{\circ }-2(∠PAB+∠PCB)+∠D=∠B$。$\because ∠P+∠PAD=∠PCD+∠D$，$∠BAD+∠B=∠BCD+∠D$，$\therefore ∠P+∠PAD-∠BAD-∠B=∠PCD-∠BCD$，$\therefore ∠P-∠PAB-∠B=∠PCB$，$\therefore ∠P-∠B=∠PAB+∠PCB$，$\therefore 180^{\circ }-2(∠P-∠B)+∠D=∠B$，即$∠P=90^{\circ }+\frac {1}{2}(∠B+∠D)$。
(2)$∠P=180^{\circ }-\frac {1}{2}(∠B+∠D)$。理由如下：连接PB，PD。
∵直线AP平分$∠BAD$的外角$∠FAD$，CP平分$∠BCD$的外角$∠BCE$，$\therefore ∠FAP=∠PAO$，$∠PCE=∠PCB$。$\because ∠APB+∠PBA+∠PAB=180^{\circ }$，$∠PCB+∠PBC+∠BPC=180^{\circ }$，$\therefore ∠APC+∠ABC+∠PCB+∠PAB=360^{\circ }$，同理得$∠APC+∠ADC+∠PCD+∠PAD=360^{\circ }$，$\therefore 2∠APC+∠ABC+∠ADC+∠PCB+∠PAB+∠PCD+∠PAD=720^{\circ }$，$\therefore 2∠APC+∠ABC+∠ADC+∠PCE+∠PAB+∠PCD+∠PAF=720^{\circ }$。$\because ∠PCE+∠PCD=180^{\circ }$，$∠PAB+∠PAF=180^{\circ }$，$\therefore 2∠APC+∠ABC+∠ADC=360^{\circ }$，$\therefore ∠APC=180^{\circ }-\frac {1}{2}(∠ABC+∠ADC)$，即$∠P=180^{\circ }-\frac {1}{2}(∠B+∠D)$。