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八年级数学解答题一般
题目
【问题初探】
(1)(1)如图11,在ABC\triangle ABC中,请说明:A+B+C=180\angle A+\angle B+\angle C=180^{\circ},为解决这一问题,同学们多用于下列两种思路:
①如图22,延长BCBCDD,过点CC作射线CECEBABA,相当于把A\angle A,B\angle B都移到了顶点CC的位置,利用图形特点获得A\angle A,B\angle B,ACB\angle ACB的数量关系;
②如图33,过点AA作直线PQPQBCBC,相当于把B\angle B,C\angle C都移到了顶点AA的位置,再利用图形特点获得BAC\angle BAC,B\angle B,C\angle C的数量关系;
请你选择上述的一种思路,说明ABC\triangle ABC的内角和为180180^{\circ}.
【类比分析】
(2)(2)如图44,已知ABC\triangle ABC,过点AA作直线PQPQBC,RBC,R为线段ABAB上一点,连接RQRQ,RCRC,若1=53\angle 1=53^{\circ},2=29\angle 2=29^{\circ},3=24\angle 3=24^{\circ},求QRC\angle QRC的度数.
【学以致用】
(3)(3)如图5,MN5,MNGTGT,MNE=12ENF∠MNE=\frac{1}{2}∠ENF,ETG=13ETF∠ETG=\frac{1}{3}∠ETF,若MNE=α\angle MNE=\alpha,E=β\angle E=\beta,F=γ\angle F=\gamma,请你判断α\alpha,β\beta,γ\gamma三者之间的数量关系,并说明理由.
知识点:等腰三角形的性质、等边三角形的性质、平行四边形的性质、平行四边形的判定、解直角三角形章节:第13章 三角形 / 13.1 三角形的概念

答案与解析

答案

(1)(1)证明:选择①,如图2\ \延长BCBCDD,过点CC作射线CECEBABA
ACE=A\therefore \angle ACE=\angle ADCE=B\angle DCE=\angle B
A+B+ACB=ACE+DCE+ACB=180\therefore \angle A+\angle B+\angle ACB=\angle ACE+\angle DCE+\angle ACB=180^{\circ}
选择②,如图33过点AABCBC的平行线PQPQ
C=CAQ\therefore \angle C=\angle CAQB=BAP\angle B=\angle BAP
BAC+B+C=BAC+BAP+CAQ=180\therefore \angle BAC+\angle B+\angle C=\angle BAC+\angle BAP+\angle CAQ=180^{\circ}
(2)(2)如图44,作RHRHBC,BC,CRH=2=29\angle CRH=\angle 2=29^{\circ}

PQ\because PQBCBC
RH\therefore RHPQPQ
ARH=1=53\therefore \angle ARH=\angle 1=53^{\circ}
3=24\because \angle 3=24^{\circ}
QRH=ARH3=5324=29\therefore \angle QRH=\angle ARH-\angle 3=53^{\circ}-24^{\circ}=29^{\circ}
QRC=QRH+CRH=29+29=58\therefore \angle QRC=\angle QRH+\angle CRH=29^{\circ}+29^{\circ}=58^{\circ}
(3)(3)γ+4βα=360\gamma +4\beta -\alpha =360^{\circ},理由如下:
如图55,延长TETEMNMN于点LL,延长TFTFMNMN于点KK

MN\because MNGTGT
ELN=ETG\therefore \angle ELN=\angle ETGGTF+FKN=180\angle GTF+\angle FKN=180^{\circ}
α=MNE=12ENF\because \alpha =\angle MNE=\frac{1}{2}\angle ENF
ENF=2α\therefore \angle ENF=2\alpha
MNF=3α\therefore \angle MNF=3\alpha
FNK=180MNF=1803α\therefore \angle FNK=180^{\circ}-\angle MNF=180^{\circ}-3\alpha
FKN=TFNFNK=γ(1803α)=γ+3α180\therefore \angle FKN=\angle TFN-\angle FNK=\gamma -\left(180^{\circ}-3\alpha \right)=\gamma +3\alpha -180^{\circ}
FTG=180FKN=180(γ+3α180)=360γ3α\therefore \angle FTG=180^{\circ}-\angle FKN=180^{\circ}-\left(\gamma +3\alpha -180^{\circ}\right)=360^{\circ}-\gamma -3\alpha
ETG=13ETF\because \angle ETG=\frac{1}{3}\angle ETF
ETG=14FTG=14(360γ3α)\therefore \angle ETG=\frac{1}{4}\angle FTG=14\left(360^{\circ}-\gamma -3\alpha \right)
ELN=NETMNE=βα\because \angle ELN=\angle NET-\angle MNE=\beta -\alpha
βα=14(360γ3α)\therefore \beta -\alpha =\frac{1}{4}(360^{\circ}-\gamma -3\alpha )
γ+4βα=360\therefore \gamma +4\beta -\alpha =360^{\circ}.

解析

(1)(1)证明:选择①,如图2\ \延长BCBCDD,过点CC作射线CECEBABA
ACE=A\therefore \angle ACE=\angle ADCE=B\angle DCE=\angle B
A+B+ACB=ACE+DCE+ACB=180\therefore \angle A+\angle B+\angle ACB=\angle ACE+\angle DCE+\angle ACB=180^{\circ}
选择②,如图33过点AABCBC的平行线PQPQ
C=CAQ\therefore \angle C=\angle CAQB=BAP\angle B=\angle BAP
BAC+B+C=BAC+BAP+CAQ=180\therefore \angle BAC+\angle B+\angle C=\angle BAC+\angle BAP+\angle CAQ=180^{\circ}
(2)(2)如图44,作RHRHBC,BC,CRH=2=29\angle CRH=\angle 2=29^{\circ}

PQ\because PQBCBC
RH\therefore RHPQPQ
ARH=1=53\therefore \angle ARH=\angle 1=53^{\circ}
3=24\because \angle 3=24^{\circ}
QRH=ARH3=5324=29\therefore \angle QRH=\angle ARH-\angle 3=53^{\circ}-24^{\circ}=29^{\circ}
QRC=QRH+CRH=29+29=58\therefore \angle QRC=\angle QRH+\angle CRH=29^{\circ}+29^{\circ}=58^{\circ}
(3)(3)γ+4βα=360\gamma +4\beta -\alpha =360^{\circ},理由如下:
如图55,延长TETEMNMN于点LL,延长TFTFMNMN于点KK

MN\because MNGTGT
ELN=ETG\therefore \angle ELN=\angle ETGGTF+FKN=180\angle GTF+\angle FKN=180^{\circ}
α=MNE=12ENF\because \alpha =\angle MNE=\frac{1}{2}\angle ENF
ENF=2α\therefore \angle ENF=2\alpha
MNF=3α\therefore \angle MNF=3\alpha
FNK=180MNF=1803α\therefore \angle FNK=180^{\circ}-\angle MNF=180^{\circ}-3\alpha
FKN=TFNFNK=γ(1803α)=γ+3α180\therefore \angle FKN=\angle TFN-\angle FNK=\gamma -\left(180^{\circ}-3\alpha \right)=\gamma +3\alpha -180^{\circ}
FTG=180FKN=180(γ+3α180)=360γ3α\therefore \angle FTG=180^{\circ}-\angle FKN=180^{\circ}-\left(\gamma +3\alpha -180^{\circ}\right)=360^{\circ}-\gamma -3\alpha
ETG=13ETF\because \angle ETG=\frac{1}{3}\angle ETF
ETG=14FTG=14(360γ3α)\therefore \angle ETG=\frac{1}{4}\angle FTG=14\left(360^{\circ}-\gamma -3\alpha \right)
ELN=NETMNE=βα\because \angle ELN=\angle NET-\angle MNE=\beta -\alpha
βα=14(360γ3α)\therefore \beta -\alpha =\frac{1}{4}(360^{\circ}-\gamma -3\alpha )
γ+4βα=360\therefore \gamma +4\beta -\alpha =360^{\circ}.

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