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初中数学八年级下册(648题)


第361题

如图,在\({\rm{Rt}}\Delta ABC\)中,\(\angle BAC = 90^\circ \),\(AD\)是边\(BC\)上的中线,过点\(A\)作\(AE//BC\),过点\(D\)作\(DE//AB\),\(DE\)与\(AC\)、\(AE\)分别交于点\(O\)、\(E\),连接\(EC\),求证:四边形\(ADCE\)是菱形。


图片 11    



参考答案:证明: \(\because AE//BC\),\(DE//AB\),

\(\therefore \)四边形\(ABDE\)为平行四边形;

\(\therefore \)\(AE = BD\),

\(\because AD\)是边\(BC\)上的中线,

\(\therefore BD = CD\),

\(\therefore AE = CD\),

\(\because AE//CD\)

\(\therefore \)四边形\(ADCE\)是平行四边形,

又\(\because \angle BAC = 90^\circ \),\(AD\)是边\(BC\)上的中线,

\(\therefore AD = \frac{1}{2}BC = CD\),

\(\therefore \)平行四边形\(ADCE\)是菱形。


第362题

如图,平行四边形\(ABCD\)中,作\(\angle DAB\)的角平分线AE,交边\(CD\)于点\(E\),过点\(E\)作\(EF//BC\)交\(AB\)于点\(F\),连接\(EF\),求证:四边形\(AFED\)是菱形。


   



参考答案:证明:\(\because \)四边形\(ABCD\)是平行四边形,

\(\therefore AB//CD\),\(AD//BC\)

\(\therefore \angle DEA = \angle FAE\),

\(\because AE\)平分\(\angle BAD\),

\(\therefore \angle DAE = \angle FAE\),

\(\therefore \angle DEA = \angle DAE\),

\(\therefore AD = ED\),

\(\because EF//BC,AD//BC\)

\(\therefore EF//AD\) ,

\(\because DE//AF\)

\(\therefore \)四边形\(AFED\)是平行四边形,

又\(\because AD = ED\),

\(\therefore \)平行四边形\(AFED\)是菱形。


第363题

如图,在等腰\(\Delta ABC\)中,\(AB = BC\),\(BO \bot AC\)于点\(O\),点\(D\)是\(BO\)上一点,延长\(BO\)至点\(E\),使\(OE = OD\),连接\(EC,EA\). 求证:四边形\(ADCE\)是菱形。


   



参考答案:证明:\(\because AB = BC\),\(BO \bot AC\),

\(\therefore AO = CO\),

又\(\because OE = OD\),

\(\therefore \)四边形\(ADCE\)是平行四边形,

\(\because DE⊥AC\),

\(\therefore \)平行四边形\(ADCE\)是菱形。


第364题

如图,在\(\Delta ABC\)中,\(\angle BAC = 90^\circ \),\(AD \bot BC\)于\(D\),\(CE\)平分\(\angle ACB\),交\(AB\)于点\(E\),交\(AD\)于点\(G\), \(EF \bot BC\)于点\(F\),连接FG,求证:四边形\(AEFG\)是菱形。


图片 22    



参考答案:证明:\(\because CE\)平分\(\angle ACB\),\(EF \bot BC\),\(\angle BAC = 90^\circ (EA \bot CA)\),

\(\therefore AE = EF\),\(∠ACE=∠BCE\)

\(\because AD \bot BC\),

\(\therefore \angle ADC = 90^\circ \),

\(\because \angle BAC = 90^\circ \),

\(\therefore \angle BCE + \angle CGD = 90^\circ \),\(\angle ACE + \angle AEC = 90^\circ \),

\(\therefore \angle CGD = \angle AEC\),

\(\because ∠CGD=∠AGE\),

\(\therefore \angle AGE = \angle AEC\)

\(\therefore AG=AE=EF\)

\(\because AD \bot BC\),\(EF \bot BC\),

\(\therefore AD//EF\),

即\(AG//EF\),\(AG = EF\),

\(\therefore \)四边形\(AEFG\)是平行四边形,

\(\because AE = EF\),

\(\therefore \)平行四边形\(AEFG\)是菱形。


第365题

如图,在平行四边形\(ABCD\)中,点\(O\)是对角线\(BD\)的中点,过点\(O\)作\(EF \bot BD\),垂足为点\(O\),且分别交边\(AD\),\(BC\)于点\(E\),\(F\).连接\(BE,DF\)。求证:四边形\(BEDF\)是菱形。


    



参考答案:证明:\(\because \)四边形\(ABCD\)是平行四边形,\(O\)为对角线\(BD\)的中点,

\(\therefore BO = DO\),\(AD//BC\),

\(\therefore \)\(\angle EDB = \angle FBO\),

\(\because \)在\(\Delta EOD\)和\(\Delta FOB\)中,

\(\left\{ {\begin{array}{*{20}{l}} {\angle EDO = \angle FBO} \\ {OD = OB} \\ {\angle EOD = \angle FOB} \end{array}} \right.\),

\(\therefore \Delta DOE \cong \Delta BOF(ASA)\);

\(\therefore OE = OF\),

又\(\because OB = OD\),

\(\therefore \)四边形\(BEDF\)是平行四边形,

\(\because EF \bot BD\),

\(\therefore \)平行四边形\(BEDF\)为菱形。


第366题

如图,平行四边形ABCD的对角线\(AC\),\(BD\)相交于点\(O\),且\(AE//BD\),\(BE//AC\),\(OE = CD\),求证:四边形\(ABCD\)是菱形。


图片 20    



参考答案:证明:
\(\because AE//BD\),\(BE//AC\),
\(\therefore \)四边形\(AEBO\)是平行四边形,
\(\because \)四边形\(ABCD\)是平行四边形,
\(\therefore DC = AB\).
\(\because OE = CD\),
\(\therefore OE = AB\).
\(\therefore \)平行四边形\(AEBO\)是矩形,
\(\therefore \angle BOA = 90^\circ \).
\(\therefore AC \bot BD\).
\(\therefore \)平行四边形\(ABCD\)是菱形。


第367题

如图,在平行四边形\(ABCD\)中,\(∠BAC=90°\),\(E\)、\(F\)分别为\(AD\)、\(BC\)的中点,点\(M\)、\(N\)在对角线\(AC\)上,且\(AM = CN\),求证:四边形\(EMFN\)是菱形。




参考答案:


证明:连接\(EF\)\(AC\)于点\(O\)



\(\because \)四边形\(ABCD\)是平行四边形,


\(\therefore AD//BC\)\(AD = BC\)


\(\therefore \angle EAM = \angle FCN\)


\(\because E\)\(F\)分别为\(AD\)\(BC\)的中点,


\(\therefore AE = \frac{1}{2}AD,{\rm{ }}CF = BF = \frac{1}{2}BC\)


\(\therefore AE=CF,AE=BF\)  


\(\because \)\(\Delta AEM\)\(\Delta CFN\)中,


\(\left\{ {\begin{array}{*{20}{l}} {AE = CF} \\ {\angle EAM = \angle FCN} \\ {AM = CN} \end{array}} \right.\)


\(\therefore \Delta AEM \cong \Delta CFN(SAS)\)


\(\therefore EM = FN\)\(∠1=∠2\)


\(\therefore ∠3=∠4\)


\(\therefore EM//FN\)


\(\therefore \)四边形\(EMFN\)是平行四边形;


\(\because \)\(AE//BF\)\(AE = BF\)


\(\therefore \)四边形\(AEFB\)是平行四边形,


\(\therefore AB//EF\)


\(\because ∠BAC=90°\)


\(\therefore \angle COF = \angle BAC = 90^\circ \)


\(\therefore EF \bot MN\)


\(\therefore \)平行四边形\(EMFN\)是菱形



第368题



参考答案:证明:
\(\because BD\)平分\(\angle ABC\),
\(\therefore \angle ABD = \angle CBD\),
\(\because \)在\(\Delta ABD\)和\(\Delta CBD\)中\(\left\{ {\begin{array}{*{20}{l}}
{\angle ABD = \angle CBD} \\
{\angle A = \angle C} \\
{BD = BD}
\end{array}} \right.\),
\(\therefore \Delta ABD≌\Delta CBD(AAS)\),
\(\therefore AB = BC\),\(AD = DC\),
\(\because AB = AD\),
\(\therefore AB = BC = DC = AD\),
\(\therefore \)四边形\(ABCD\)是菱形。


第369题



参考答案:证明:\(\because AC\)是\(BD\)的垂直平分线,点\(F\)在\(AC\)上,

\(\therefore AB = AD\),\(BC = CD\),\(FD=FB\),

\(\because \)在\(\Delta ABF\)和\(\Delta ADF\)中,

\(\left\{ {\begin{array}{*{20}{l}} {AB = AD} \\ {FB = FD} \\ {AF = AF} \end{array}} \right.\),

\(\therefore \Delta ABF \cong \Delta ADF({\rm{SSS}})\),

\(\therefore \angle ABF = \angle ADF\),\(∠BAC=∠DAC\)

\(\because \angle BEC = \angle ADF\),

\(\therefore \angle BEC = \angle ABF\),

\(\therefore AB//CD\),

\(\therefore \angle BAC = \angle ACD\),

\(\therefore \angle DAC = \angle ACD\),

\(\therefore AD = CD\),

\(\because AB = AD\),\(BC = CD\),

\(\therefore AB = CB = CD = AD\),

\(\therefore \)四边形\(ABCD\)是菱形。


第370题


A.\(120^\circ \)

B.\(135^\circ \)

C.\(145^\circ \)

D.\(150^\circ \)


参考答案:D


第371题


A.\(\sqrt 2 \)

B.4

C.\(2\sqrt 2 \)

D.\(\frac{{\sqrt 2 }}{2}\)


参考答案:C




第374题


A.\(9\sqrt 2 \)

B.18

C.24

D.36


参考答案:B



第376题


A.\(15^\circ \)

B.\(32.5^\circ \)

C.\(22.5^\circ \)

D.\(30^\circ \)


参考答案:C



第378题

如图,延长正方形\(ABCD\)的边\(BA\)至点\(E\),使\(AE = BD\),连接\(CE\),则\(\angle E\)为(   )。


   


A.\(22.5^\circ \)

B.\(25^\circ \)

C.\(30^\circ \)

D.\(45^\circ \)


参考答案:A




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