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初中数学八年级上册试题库(555题)



第122题

如图,点\(O\)\(\Delta ABC\)内,且到三边的距离相等,\(\angle A = 64^\circ \),则\(\angle BOC\)的度数为\((\)  \()\)


A.\(58^\circ \)

B.\(64^\circ \)

C.\(122^\circ \)

D.\(124^\circ \)


参考答案:C




第125题

如图,\(AD\)\(\angle BAC\)的平分线,\(DF \bot AC\)\(F\)\(\angle B = 90^\circ \)\(DE = DC\),试说明:\(BE = CF\)



参考答案:\(\because \angle B = 90^\circ \),\(\therefore BD \bot AB\)。\(\because AD\)为\(\angle BAC\)的平分线,且\(DF \bot AC\),\(\therefore DB = DF\)。在\({\rm{Rt}}\Delta {\rm{BDE}}\)和\({\rm{Rt}}\Delta {\rm{FDC}}\)中,\(\left\{ {\begin{array}{*{20}{l}}
{DE = DC} \\
{DB = DF}
\end{array}} \right.\),\(\therefore {\rm{Rt}}\Delta {\rm{BDE}} \cong {\rm{Rt}}\Delta {\rm{FDC}}({\rm{HL}})\),\(\therefore BE = CF\)。


第126题

已知,如图,\(\Delta ABC\)中,\(\angle C = 90^\circ \)\(D\)\(AB\)上一点,\(DE \bot CD\)\(D\),交\(BC\)\(E\),且有\(AC = AD = CE\),求证:\(DE = \frac{1}{2}CD\)



参考答案:

证明:如图,过点\(A\)\(AF \bot CD\)

\(\because \angle C = 90^\circ \)\(DE \bot CD\)

\(\therefore \angle ACF + \angle DCE = \angle DCE + \angle DEC\)

\(\therefore \angle ACF = \angle DEC\)

\(\Delta ACF\)\(\Delta CED\)中,

\(\left\{ {\begin{array}{*{20}{l}}
{\angle ACF = \angle DEC} \\
{\angle AFC = \angle CDE} \\
{AC = CE}
\end{array}} \right.\)

\(\therefore \Delta ACF \cong \Delta CED(AAS)\)

\(\therefore CF = DE\)

\(\because AC = AD\),且\(AF \bot CD\)

\(\therefore CF = \frac{1}{2}CD\)

\(\therefore DE = \frac{1}{2}CD\)




第129题

如图,点\(B\)\(E\)\(C\)\(F\)在同一条直线上,\(AB//DE\)\(AB = DE\),要使\(\Delta ABC \cong \Delta DEF\),添加的一个条件可以是___。

图片 5



参考答案:①\(BC = EF\)(\(BE = CF\))②\(\angle A = \angle D\)③\(\angle ACB = \angle F\)(\(AC//DF\))


第130题

如图,\(BD\)平分\(\angle ABC\)\(F\)\(G\)分别是\(BA\)\(BC\)上的点\((BF \ne BG)\)\(EF = EG\),则\(\angle BFE\)\(\angle BGE\)的数量关系为\((\)  \()\)

图片 2


A.\(\angle BFE + \angle BGE = 90^\circ \)

B.\(\angle BFE + \angle BGE = 180^\circ \)

C.\(\angle BFE = 2\angle BGE\)

D.\(\angle BFE - \angle BGE = 90^\circ \)


参考答案:B


第131题

如图,已知\(AB//CD\)\(OA = OD\)\(AE = DF\).试说明:\( EB=CF\)



参考答案:证明:\(\because AB//CD\),\(\therefore \angle 3 = \angle 4\),在\(\Delta ABO\)和\(\Delta DCO\)中,\(\left\{ {\begin{array}{*{20}{l}}
{\angle 3 = \angle 4} \\
{OA = OD} \\
{\angle 1 = \angle 2}
\end{array}} \right.\),\(\therefore \Delta ABO \cong \Delta DCO(ASA)\),\(\therefore AB = CD\),\(\because \angle 3 = \angle 4\),\(\therefore \angle CDF = \angle BAE\),在\(\Delta ABE\)和\(\Delta DCF\)中,\(\left\{ {\begin{array}{*{20}{l}}
{AE = DF} \\
{\angle BAE = \angle CDF} \\
{AB = CD}
\end{array}} \right.\),\(\therefore \Delta ABE \cong \Delta DCF(SAS)\),\( EB=CF\)。


第132题

求证:\(\Delta ABM \cong \Delta BCN\)



参考答案:证明:正五边形\(ABCDE\)中,\(AB = BC\),\(\angle ABM = \angle C\),在\(\Delta ABM\)和\(\Delta BCN\)中\(\left\{ {\begin{array}{*{20}{l}}
{AB = BC} \\
{\angle ABM = \angle C} \\
{BM = CN}
\end{array}} \right.\),\(\therefore \Delta ABM \cong \Delta BCN(SAS)\);


第133题

\(\angle APN\)的度数。



参考答案:\(\because \Delta ABM \cong \Delta BCN\),\(\therefore \angle BAM = \angle CBN\),\(\therefore \angle APN = \angle BAM + \angle ABP\)\( = \angle CBN + \angle ABP\)\( = \angle ABC\)\( = \frac{{(5 - 2) \times 180^\circ }}{5}\)\( = 108^\circ \)。


第134题

如图,\(\Delta ABC\)中,\(D\)\(BC\)中点,\(E\)\(F\)分别为\(AB\)\(AC\)上的点,且满足\(ED \bot DF\),连接\(EF\),求证:\(BE + FC > EF\)

图片 7



参考答案:

证明:延长\(FD\)\(M\)使\(MD = FD\),连接\(BM\)\(EM\)

\(\Delta BDM\)\(\Delta CDF\)

\(\begin{cases}
BD = CD \\
\angle BDM = \angle FDC \\
MD = FD \\
\end{cases}\)

\(\therefore \Delta BDM \cong \Delta CDF(SAS)\)

\(\therefore BM = CF\)

\(\because ED \bot DF\)\(MD = FD\)

\(\therefore EM = EF\)

\(\Delta BME\)中,\(BE + BM > EM\)

\(\therefore BE + FC > EF\)

图片 6






第139题

如图,在\(\Delta ABC\)\(\Delta ADE\)中,\(AB = AE\)\(AB//ED\)\(\angle DAB = \angle BAE + \angle B\)

求证:\(AC = ED\)

图片 5



参考答案:证明:\(\because AB//ED\),\(\therefore \angle EAB = \angle E\),\(\because \angle DAB = \angle BAE + \angle B\),\(\angle DAB = \angle BAE + \angle DAE\),\(\therefore \angle B = \angle DAE\),在\(\Delta ABC\)和\(\Delta EAD\)中,\(\left\{ {\begin{array}{*{20}{l}}
{\angle B = \angle DAE} \\
{AB = EA} \\
{\angle BAC = \angle E}
\end{array}} \right.\),\(\therefore \Delta ABC \cong \Delta EAD(ASA)\),\(\therefore AC = ED\).


第140题

如图,已知\(AC\)平分\(\angle BAD\)\(CE \bot AB\)\(E\)\(CF \bot AD\)\(F\),且\(BC = CD\).求证:

图片 3

(1)\(\Delta BCE \cong \Delta DCF\)

(2)\(AB + AD = 2AE\)



参考答案:证明:(1)\(\because AC\)是角平分线,\(CE \bot AB\)于\(E\),\(CF \bot AD\)于\(F\),\(\therefore CE = CF\),在\({\rm{Rt}}\Delta {\rm{BCE}}\)和\({\rm{Rt}}\Delta {\rm{DCF}}\)中,\(\left\{ {\begin{array}{*{20}{l}}
{BC = DC} \\
{CE = CF}
\end{array}} \right.\)\(\therefore \Delta BCE \cong \Delta DCF(HL)\);(2)在\({\rm{Rt}}\Delta {\rm{FAC}}\)和\({\rm{Rt}}\Delta {\rm{EAC}}\)中,\(\left\{ {\begin{array}{*{20}{l}}
{AC = AC} \\
{CE = CF}
\end{array}} \right.\),\(\therefore Rt\Delta FAC \cong Rt\Delta EAC(HL)\),\(\therefore AF = AE\),\(\because \Delta BCE \cong \Delta DCF\),\(\therefore BE = DF\),\(\therefore AB + AD = (AE + BE) + (AF - DF) = 2AE\)。


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