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1. 如图,在$Rt△ABC$中,D 是边 AC 的中点,$∠BDC= 60^{\circ },AC= 6$,则边 BC 的长是(
A.3
B.6
C.$\sqrt {3}$
D.$3\sqrt {3}$
A
)A.3
B.6
C.$\sqrt {3}$
D.$3\sqrt {3}$
答案
A [解析]∵在$Rt\triangle ABC$中,$∠ABC=90^{\circ }$,D 是边AC 的中点,
$\therefore BD=\frac {1}{2}AC=CD.$
$\because ∠BDC=60^{\circ },$
$\therefore \triangle BDC$为等边三角形,
$\therefore BC=CD=\frac {1}{2}AC=\frac {1}{2}×6=3.$
$\therefore BD=\frac {1}{2}AC=CD.$
$\because ∠BDC=60^{\circ },$
$\therefore \triangle BDC$为等边三角形,
$\therefore BC=CD=\frac {1}{2}AC=\frac {1}{2}×6=3.$
2. 如图,直线$l// m$,等边三角形 ABC 的两个顶点 B,C 分别落在直线 l,m 上.若$∠ABE= 21^{\circ }$,则$∠ACD$的度数是(
A.$45^{\circ }$
B.$39^{\circ }$
C.$29^{\circ }$
D.$21^{\circ }$
B
)A.$45^{\circ }$
B.$39^{\circ }$
C.$29^{\circ }$
D.$21^{\circ }$
答案
B [解析]∵直线$l// m,$
$\therefore ∠EBC+∠DCB=180^{\circ },$
即$∠EBA+∠ABC+∠ACB+∠ACD=180^{\circ }.$
$\because \triangle ABC$是等边三角形,
$\therefore ∠ABC=∠ACB=60^{\circ }.$
又$\because ∠ABE=21^{\circ },$
$\therefore 21^{\circ }+60^{\circ }+60^{\circ }+∠ACD=180^{\circ },$
$\therefore ∠ACD=39^{\circ }.$
$\therefore ∠EBC+∠DCB=180^{\circ },$
即$∠EBA+∠ABC+∠ACB+∠ACD=180^{\circ }.$
$\because \triangle ABC$是等边三角形,
$\therefore ∠ABC=∠ACB=60^{\circ }.$
又$\because ∠ABE=21^{\circ },$
$\therefore 21^{\circ }+60^{\circ }+60^{\circ }+∠ACD=180^{\circ },$
$\therefore ∠ACD=39^{\circ }.$
3. 如图,在$△ABC$中,$AB= AC= 10,BC= 8$,AD 平分$∠BAC$交 BC 于点 D,E 为 AC 的中点,连接 DE,则$△CDE$的周长为____
14
.答案
14 [解析]∵$AB=AC$,AD 平分$∠BAC,BC=8,$
$\therefore AD⊥BC,CD=BD=\frac {1}{2}BC=4.$
∵E 为 AC 的中点,
$\therefore DE=CE=\frac {1}{2}AC=5,$
$\therefore \triangle CDE$的周长为$CD+DE+CE=4+5+5=14.$
$\therefore AD⊥BC,CD=BD=\frac {1}{2}BC=4.$
∵E 为 AC 的中点,
$\therefore DE=CE=\frac {1}{2}AC=5,$
$\therefore \triangle CDE$的周长为$CD+DE+CE=4+5+5=14.$
4. 如图,D,E 分别是等边三角形 ABC 的边 BC,AC 上的点,且$BD= CE$,BE 与 AD 交于点 F.求证：$AD= BE.$
证明:∵$\triangle ABC$是等边三角形,
$\therefore AB=BC,∠ABD=∠BCE=60^{\circ }.$
在$\triangle ABD$和$\triangle BCE$中,
$\left\{\begin{array}{l} AB=BC,\\ ∠ABD=∠BCE,\\ BD=CE,\end{array}\right.$
$\therefore \triangle ABD\cong \triangle BCE$(
$\therefore AD=BE.$
证明:∵$\triangle ABC$是等边三角形,
$\therefore AB=BC,∠ABD=∠BCE=60^{\circ }.$
在$\triangle ABD$和$\triangle BCE$中,
$\left\{\begin{array}{l} AB=BC,\\ ∠ABD=∠BCE,\\ BD=CE,\end{array}\right.$
$\therefore \triangle ABD\cong \triangle BCE$(
SAS
),$\therefore AD=BE.$
答案
证明:∵$\triangle ABC$是等边三角形,
$\therefore AB=BC,∠ABD=∠BCE=60^{\circ }.$
在$\triangle ABD$和$\triangle BCE$中,
$\left\{\begin{array}{l} AB=BC,\\ ∠ABD=∠BCE,\\ BD=CE,\end{array}\right.$
$\therefore \triangle ABD\cong \triangle BCE(SAS),$
$\therefore AD=BE.$
$\therefore AB=BC,∠ABD=∠BCE=60^{\circ }.$
在$\triangle ABD$和$\triangle BCE$中,
$\left\{\begin{array}{l} AB=BC,\\ ∠ABD=∠BCE,\\ BD=CE,\end{array}\right.$
$\therefore \triangle ABD\cong \triangle BCE(SAS),$
$\therefore AD=BE.$
5. (江苏连云港期末)如图,将边长为 3 个单位长度的等边三角形 ABC 沿边 BC 向右平移 2 个单位长度得到$△DEF$,连接 AD,则四边形 ABFD 的周长为(
A.13
B.14
C.15
D.16
13
)A.13
B.14
C.15
D.16
答案
A [解析]∵将边长为 3 个单位长度的等边三角形ABC 沿边 BC 向右平移 2 个单位长度得到$\triangle DEF,$
$\therefore DF=AC=BC=AB=3,AD=CF=2,$
$\therefore$ 四边形 ABFD 的周长为$AB+BC+CF+DF+AD=3+3+2+3+2=13.$
$\therefore DF=AC=BC=AB=3,AD=CF=2,$
$\therefore$ 四边形 ABFD 的周长为$AB+BC+CF+DF+AD=3+3+2+3+2=13.$
6. (内蒙古呼伦贝尔中考)如图,在$△ABC$中,$∠C= 90^{\circ },∠B= 30^{\circ }$,以点 A 为圆心,适当长为半径画弧,分别交 AB,AC 于点 M,N,再分别以点 M,N 为圆心,大于$\frac {1}{2}MN$的长为半径画弧,两弧交于点 P,连接 AP 并延长交 BC 于点 D.若$△ACD$的面积为 8,则$△ABD$的面积为(
A.8
B.16
C.12
D.24
B
)A.8
B.16
C.12
D.24
答案
B [解析]∵$∠C=90^{\circ },∠B=30^{\circ },$
$\therefore ∠BAC=60^{\circ }.$
由题意,知 AD 平分$∠BAC,$
$\therefore ∠CAD=∠DAB=30^{\circ },$
$\therefore CD=\frac {1}{2}AD,∠B=∠BAD,$
$\therefore AD=BD,$
$\therefore CD=\frac {1}{2}BD,$
$\therefore \frac {S_{\triangle ACD}}{S_{\triangle ABD}}=\frac {\frac {1}{2}CD\cdot AC}{\frac {1}{2}BD\cdot AC}=\frac {CD}{BD}=\frac {1}{2}.$
又$\because \triangle ACD$的面积为 8,
$\therefore \triangle ABD$的面积为$2×8=16.$
$\therefore ∠BAC=60^{\circ }.$
由题意,知 AD 平分$∠BAC,$
$\therefore ∠CAD=∠DAB=30^{\circ },$
$\therefore CD=\frac {1}{2}AD,∠B=∠BAD,$
$\therefore AD=BD,$
$\therefore CD=\frac {1}{2}BD,$
$\therefore \frac {S_{\triangle ACD}}{S_{\triangle ABD}}=\frac {\frac {1}{2}CD\cdot AC}{\frac {1}{2}BD\cdot AC}=\frac {CD}{BD}=\frac {1}{2}.$
又$\because \triangle ACD$的面积为 8,
$\therefore \triangle ABD$的面积为$2×8=16.$
7. 练思维·推理能力 如图,$∠MON= 30^{\circ }$,点$A_{1},A_{2},A_{3}... $在射线 ON 上,点$B_{1},B_{2},B_{3}... $在射线 OM 上,$△A_{1}B_{1}A_{2},△A_{2}B_{2}A_{3},△A_{3}B_{3}A_{4}... $均为等边三角形.以此类推,若$OA_{1}= 1$,则$△A_{2023}B_{2023}A_{2024}$的边长为(
A.2024
B.4042
C.$2^{2023}$
D.$2^{2022}$
$2^{2022}$
)A.2024
B.4042
C.$2^{2023}$
D.$2^{2022}$
答案
D [解析]∵$\triangle A_{1}B_{1}A_{2}$为等边三角形,
$\therefore ∠B_{1}A_{1}A_{2}=60^{\circ },$
$\therefore ∠OB_{1}A_{1}=∠B_{1}A_{1}A_{2}-∠MON=30^{\circ },$
$\therefore ∠OB_{1}A_{1}=∠MON,$
$\therefore A_{1}B_{1}=OA_{1}=1.$
同理可得,$A_{2}B_{2}=OA_{2}=2,A_{3}B_{3}=OA_{3}=2^{2},$
$\therefore \triangle A_{2023}B_{2023}A_{2024}$的边长为$2^{2022}.$
$\therefore ∠B_{1}A_{1}A_{2}=60^{\circ },$
$\therefore ∠OB_{1}A_{1}=∠B_{1}A_{1}A_{2}-∠MON=30^{\circ },$
$\therefore ∠OB_{1}A_{1}=∠MON,$
$\therefore A_{1}B_{1}=OA_{1}=1.$
同理可得,$A_{2}B_{2}=OA_{2}=2,A_{3}B_{3}=OA_{3}=2^{2},$
$\therefore \triangle A_{2023}B_{2023}A_{2024}$的边长为$2^{2022}.$
8. (江苏泰州)如图,已知直线$l_{1}// l_{2}$,将等边三角形按如图所示的方式放置.若$∠α=40^{\circ }$,则$∠β=$____
20°
.答案
$20^{\circ }$ [解析]如图,过点 A 作$AD// l_{1}$.根据平行线的性质可得,$∠BAD=∠β.\because l_{1}// l_{2},\therefore AD// l_{2},$$\therefore ∠DAC=∠α=40^{\circ }$.根据等边三角形的性质可得,$∠BAC=60^{\circ },\therefore ∠β=∠BAD=∠BAC-∠DAC=60^{\circ }-40^{\circ }=20^{\circ }.$
