《2025年拔尖特训八年级数学上册人教版》

1. (2023·金昌)如图,BD是等边三角形ABC的边AC上的高,以点D为圆心,DB长为半径画弧,交BC的延长线于点E,连接DE,则∠DEC的度数为 (C)

A. $20^{\circ}$
B. $25^{\circ}$
C. $30^{\circ}$
D. $35^{\circ}$

答案： C

2. 如图,$AB = AC$,D是BC的中点,AB平分$∠DAE$,$AE⊥BE$,垂足为E.若$BE// AC$,则$∠C$的度数为$60^{\circ}$.

答案： $60^{\circ}$

3. (2024·济南槐荫期末)如图,$△ABC$为等边三角形,BD平分$∠ABC$交AC于点D,$DE// BC$交AB于点E.求证:
(1)$△ADE$是等边三角形.
(2)$AE=\frac{1}{2}AB$.

答案：
(1) $\because \triangle ABC$为等边三角形，
$\therefore \angle A=\angle ABC=\angle C=60^{\circ}$.
$\because DE// BC$，
$\therefore \angle AED=\angle ABC=60^{\circ}$，$\angle ADE=\angle C=60^{\circ}$.
$\therefore \angle A=\angle AED=\angle ADE$.
$\therefore \triangle ADE$是等边三角形.
(2) $\because \triangle ABC$为等边三角形，
$\therefore AB=BC=AC$.
$\because BD$平分$\angle ABC$，
$\therefore AD=\frac{1}{2}AC$.
由
(1)知，$\triangle ADE$是等边三角形，
$\therefore AE=AD$.
$\therefore AE=\frac{1}{2}AB$.

4. 如图,在等边三角形ABC中,D,E分别为AB,AC上的动点,$BD = 2AE$,连接DE,以DE为边在$△ABC$内作等边三角形DEF,连接CF.当点D从点A出发向点B运动(不与点B重合)时,$∠ECF$的度数的变化情况是 ( )

A. 不变
B. 变小
C. 变大
D. 先变大后变小

答案：
A 解析：如图，在$AC$上截取$CN=AE$，连接$FN$. $\because \triangle ABC$是等边三角形，$\therefore \angle A=60^{\circ}$，$AB=AC$.
$\because BD=2AE$，$\therefore$易得$AD=NE$.
$\because \triangle DEF$是等边三角形，$\therefore DE=EF$，$\angle DEF=60^{\circ}$. $\because \angle ADE=180^{\circ}-\angle A-\angle AED=180^{\circ}-60^{\circ}-\angle AED=120^{\circ}-\angle AED$，$\angle NEF=180^{\circ}-\angle DEF-\angle AED=180^{\circ}-60^{\circ}-\angle AED=120^{\circ}-\angle AED$，
$\therefore \angle ADE=\angle NEF$. 在$\triangle ADE$和$\triangle NEF$中，$\begin{cases}AD=NE,\\\angle ADE=\angle NEF,\\DE=EF,\end{cases}$
$\therefore \triangle ADE\cong \triangle NEF$. $\therefore AE=NF$，$\angle A=\angle FNE=60^{\circ}$. $\therefore NF=CN$.
$\therefore \angle NCF=\angle NFC$. $\because \angle FNE=\angle NCF+\angle NFC=60^{\circ}$，$\therefore \angle NCF=30^{\circ}$，即$\angle ECF=30^{\circ}$. $\therefore$当点$D$从点$A$出发向点$B$运动(不与点$B$重合)时，$\angle ECF$的度数不变.
　　　　

5. 如图,$∠ABC = 120^{\circ}$,BD平分$∠ABC$,$∠DAC = 60^{\circ}$.若$AB = 2$,$BC = 3$,则BD的长为 ( )

A. 5
B. 7
C. 8
D. 9

答案：
A 解析：如图，在$CB$的延长线上取点$E$，使$BE=AB$，连接$AE$.
$\because \angle ABC=120^{\circ}$，$\therefore \angle ABE=180^{\circ}-\angle ABC=60^{\circ}$. $\because BE=AB$，
$\therefore \triangle ABE$是等边三角形. $\therefore AE=AB$，$\angle BAE=\angle E=60^{\circ}$. $\because \angle DAC=60^{\circ}$，$\therefore \angle DAC=\angle BAE$. $\because \angle BAD=\angle BAC+\angle DAC$，$\angle EAC=\angle BAC+\angle BAE$，$\therefore \angle BAD=\angle EAC$. $\because BD$平分$\angle ABC$，$\therefore \angle ABD=\frac{1}{2}\angle ABC=60^{\circ}$. $\therefore \angle ABD=\angle E$. 在$\triangle ABD$和$\triangle AEC$中，$\begin{cases}\angle BAD=\angle EAC,\\AB=AE,\\\angle ABD=\angle E,\end{cases}$
$\therefore \triangle ABD\cong \triangle AEC$. $\therefore BD=EC$.
$\because EC=BE+BC=AB+BC=2+3=5$，$\therefore BD=5$.
　　　

6. 如图,在$△ABC$中,$∠ACB = 90^{\circ}$,$∠BAC = 30^{\circ}$,$BC = 4$,且E为边BC的中点,连接AE,以AE为边向上作等边三角形ADE,连接BD,则BD的长为______.

答案：
6 解析：如图，延长$BC$到点$T$，使得$CT=CB$，连接$AT$. 由题意，得$AC\perp BT$. 又$\because CB=CT$，$\therefore AB=AT$. $\therefore$易得$\angle BAC=\angle TAC=30^{\circ}$.
$\therefore \angle BAT=60^{\circ}$. $\therefore \triangle ABT$是等边三角形. $\because \triangle ADE$是等边三角形，
$\therefore \angle EAD=60^{\circ}$，$AD=AE$. $\therefore \angle EAD=\angle BAT$. $\therefore$易得$\angle BAD=\angle TAE$. 在$\triangle BAD$和$\triangle TAE$中，$\begin{cases}AB=AT,\\\angle BAD=\angle TAE,\\AD=AE,\end{cases}$
$\therefore \triangle BAD\cong \triangle TAE$. $\therefore BD=TE$. 由题意，易得$BC=CT=4$，$EC=EB=2$. $\therefore TE=EC+CT=6$. $\therefore BD=TE=6$. $\therefore BD$的长为6.
　　　　

7. 如图①,$△ABC$是等边三角形,点D在$△ABC$内部,且$∠BDC = 120^{\circ}$.
(1)设$∠ABD = α$,求$∠ACD$的度数(用含α的式子表示).
(2)如图②,E是BC的中点,连接AD,DE,用等式表示线段AD与DE之间的数量关系,并加以证明.

答案：
(1) $\because \triangle ABC$是等边三角形，
$\therefore \angle ABC=\angle ACB=60^{\circ}$，即$\angle ABD+\angle DBC=60^{\circ}$，$\angle ACD+\angle BCD=60^{\circ}$.
$\because \angle BDC=120^{\circ}$，
$\therefore \angle DBC+\angle BCD=180^{\circ}-120^{\circ}=60^{\circ}$.
$\therefore$易得$\angle ABD+\angle ACD=60^{\circ}$.
$\therefore \angle ACD=60^{\circ}-\angle ABD=60^{\circ}-\alpha$.
(2) $AD=2DE$.
延长$CD$至点$F$，使$DF=BD$，连接$BF$，$AF$.
$\because \angle BDC=120^{\circ}$，
$\therefore \angle BDF=180^{\circ}-\angle BDC=60^{\circ}$.
$\because DF=BD$，
$\therefore \triangle BDF$是等边三角形.
$\therefore BF=DF=BD$，$\angle FBD=\angle BFD=60^{\circ}$.
$\therefore \angle ABF+\angle ABD=60^{\circ}$.
$\because \angle ABC=\angle ABD+\angle DBC=60^{\circ}$，
$\therefore \angle ABF=\angle DBC$.
$\because \triangle ABC$是等边三角形，
$\therefore AB=BC$.
$\therefore \triangle ABF\cong \triangle CBD$.
$\therefore AF=CD$，$\angle BFA=\angle BDC=120^{\circ}$.
$\therefore \angle AFD=\angle BFA-\angle BFD=60^{\circ}$.
延长$DE$至点$G$，使$GE=DE$，连接$BG$.
$\because E$是$BC$的中点，
$\therefore BE=CE$.
又$\because \angle BEG=\angle CED$，$GE=DE$，
$\therefore \triangle BEG\cong \triangle CED$.
$\therefore BG=CD$，$\angle GBE=\angle DCE$.
$\therefore BG=AF$.
$\because \angle BDC=120^{\circ}$，
$\therefore \angle DCB+\angle DBC=60^{\circ}$.
$\therefore \angle GBE+\angle DBC=60^{\circ}$，即$\angle GBD=60^{\circ}$.
$\therefore \angle GBD=\angle AFD=60^{\circ}$.
在$\triangle DFA$和$\triangle DBG$中，
$\begin{cases}DF=DB,\\\angle AFD=\angle GBD,\\AF=GB,\end{cases}$
$\therefore \triangle DFA\cong \triangle DBG$.
$\therefore AD=GD=DE+GE=2DE$.