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1. (2024·兰州中考)如图,在$\triangle ABC$中,$AB = AC$,$\angle BAC = 130^{\circ}$,$DA \perp AC$,则$\angle ADB = $()
A. $100^{\circ}$
B. $115^{\circ}$
C. $130^{\circ}$
D. $145^{\circ}$
A. $100^{\circ}$
B. $115^{\circ}$
C. $130^{\circ}$
D. $145^{\circ}$
答案
B
2. (湖州中考)如图,$AD$,$CE分别是\triangle ABC$的中线和角平分线.若$AB = AC$,$\angle CAD = 20^{\circ}$,则$\angle ACE$的度数是()
A. $20^{\circ}$
B. $35^{\circ}$
C. $40^{\circ}$
D. $70^{\circ}$
A. $20^{\circ}$
B. $35^{\circ}$
C. $40^{\circ}$
D. $70^{\circ}$
答案
B
3. (2024·保山期中)如图,在$\text{Rt}\triangle ABC$中,$\angle C = 90^{\circ}$.分别以$A$,$B$为圆心,大于$\frac{1}{2}AB$的长为半径画弧,分别交于$M$,$N$两点,连接$MN交AC于点P$.若点$P到AB$,$BC$的距离相等,则$\angle A$的度数为()
A. $20^{\circ}$
B. $25^{\circ}$
C. $30^{\circ}$
D. $40^{\circ}$
A. $20^{\circ}$
B. $25^{\circ}$
C. $30^{\circ}$
D. $40^{\circ}$
答案
C
4. (1)(滨州中考)在等腰$\triangle ABC$中,$AB = AC$,$\angle B = 50^{\circ}$,则$\angle A$的大小为______$^{\circ}$.
(2)(云南中考)已知$\triangle ABC$是等腰三角形.若$\angle A = 40^{\circ}$,则$\triangle ABC$的顶角度数是______.
(2)(云南中考)已知$\triangle ABC$是等腰三角形.若$\angle A = 40^{\circ}$,则$\triangle ABC$的顶角度数是______.
答案
(1)80 (2)40°或100°
5. (2025·吕梁期末)在$\triangle ABC$中,$AB = AC$,$BD \perp AC于点D$,点$E在AB$边上,$CB = CE$,$CD = 5$,则$BE = $______.
答案
10 解析:如图,过点C作CF⊥AB,交AB于点F,∵BD⊥AC,CF⊥AB,∴∠BDC=∠CFB=90°.∵AB=AC,∴∠ABC=∠ACB.又∵BC=CB,∴△BDC≌△CFB(AAS),∴BF=CD=5.∵CB=CE,∴BF=EF=5,则BE=BF+EF=10.
6. 如图,在$\triangle ABC$中,$AB的垂直平分线EF交BC于点E$,交$AB于点F$,$D为线段CE$的中点,$BE = AC$.
(1)求证:$AD \perp BC$;
(2)若$\angle BAC = 75^{\circ}$,求$\angle B$的度数.
(1)求证:$AD \perp BC$;
(2)若$\angle BAC = 75^{\circ}$,求$\angle B$的度数.
答案
(1)连接AE,∵EF垂直平分AB,∴AE=BE.∵BE=AC,∴AE=AC.∵D为线段CE的中点,∴AD⊥BC.
(2)设∠B=x°,∵AE=BE,∴∠BAE=∠B=x°,由三角形外角的性质可知,∠AEC=2x°.∵AE=AC,∴∠C=∠AEC=2x°.在△ABC中,3x°+75°=180°,解得x°=35°,∴∠B=35°.
(2)设∠B=x°,∵AE=BE,∴∠BAE=∠B=x°,由三角形外角的性质可知,∠AEC=2x°.∵AE=AC,∴∠C=∠AEC=2x°.在△ABC中,3x°+75°=180°,解得x°=35°,∴∠B=35°.
7. (苏州中考)如图,在$\triangle ABC$中,$\angle BAC = 108^{\circ}$,将$\triangle ABC绕点A按逆时针方向旋转得到\triangle AB'C'$.若点$B'恰好落在BC$边上,且$AB' = CB'$,则$\angle C'$的度数为()
A. $18^{\circ}$
B. $20^{\circ}$
C. $24^{\circ}$
D. $28^{\circ}$
A. $18^{\circ}$
B. $20^{\circ}$
C. $24^{\circ}$
D. $28^{\circ}$
答案
7.C 解析:∵AB'=CB',∴∠CAB'=∠C.∵△ABC绕点A按逆时针方向旋转得到△AB'C',∴∠C'=∠C,AB'=AB,∴∠B=∠AB'B=∠CAB'+∠C=∠C+∠C=2∠C;∵∠B+∠C=180°−∠BAC=180°−108°=72°,∴3∠C=72°,∴∠C'=∠C=24°.故选C.
8. (2024·深圳期末)如图,在$\triangle ABC$中,$AB = AC$,$\angle BAC = 54^{\circ}$,点$D为AB$中点,且$OD \perp AB$,$\angle BAC的平分线与AB的垂直平分线交于点O$,将$\angle C沿EF$($E在BC$上,$F在AC$上)折叠,点$C与点O$恰好重合,则$\angle OEC$的度数为()
A. $108^{\circ}$
B. $126^{\circ}$
C. $144^{\circ}$
D. 无法确定
A. $108^{\circ}$
B. $126^{\circ}$
C. $144^{\circ}$
D. 无法确定
答案
8.A 解析:如图,连接OB,OC,∵∠BAC=54°,AO为∠BAC的平分线,∴∠BAO=$\frac{1}{2}$∠BAC=$\frac{1}{2}$×54°=27°.又∵AB=AC,∴∠ABC=$\frac{1}{2}$(180°−∠BAC)=$\frac{1}{2}$×(180°−54°)=63°.∵DO是AB的垂直平分线,∴OA=OB,∴∠ABO=∠BAO=27°,∴∠OBC=∠ABC−∠ABO=63°−27°=36°.∵AO为∠BAC的平分线,AB=AC,∴△AOB≌△AOC,∴OB=OC,∴∠OCB=∠OBC=36°,由折叠可知OE=CE,∴∠COE=∠OCB=36°,在△OCE中,∠OEC=180°−∠COE−∠OCB=180°−36°−36°=108°.故选A.
