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            • 1. 已知直线\(l_{1}/\!/l_{2}/\!/l_{3}/\!/l_{4}\),相邻的两条平行直线间的距离均为\(h\),矩形\(ABCD\)的四个顶点分别在这四条直线上,放置方式如图所示,\(AB=4\),\(BC=6\),则\(\tan α\)的值等于\((\)  \()\)
              A.\( \dfrac {2}{3}\)
              B.\( \dfrac {3}{4}\)
              C.\( \dfrac {4}{3}\)
              D.\( \dfrac {3}{2}\)
            • 2.

              如图,在\(□\)\(ABCD\)中,\(AC\)与\(BD\)相交于点\(O\),\(E\)为\(OD\)的中点,连接\(AE\)并延长交\(DC\)于点\(F\),则\(DF:FC=\)  \((\)    \()\)

              A.\(1:4\)
              B.\(1:3\)
              C.\(2:3\)
              D.\(1:2\)
            • 3. 如图,在\(\triangle ABC\)中,\(∠A=60^{\circ}\),\(BM⊥AC\)于点\(M\),\(CN⊥AB\)于点\(N\),\(P\)为\(BC\)边的中点,连接\(PM\),\(PN\),\(MN\),有下列结论:\(①PM=PN\);\(②\dfrac{AM}{AB}=\dfrac{AN}{AC}\);\(③\triangle PMN\)为等边三角形;\(④\)当\(∠ABC=45^{\circ}\)时,\(BN=\sqrt{2}PC.\)其中正确结论的个数是  \((\)    \()\)

              A.\(1\)
              B.\(2\)
              C.\(3\)
              D.\(4\)
            • 4.
              如图,在\(\triangle ABC\)中,\(DE/\!/BC\),\( \dfrac {AD}{BD}= \dfrac {1}{2}\),\(DE=4cm\),则\(BC\)的长为\((\)  \()\)
              A.\(8cm\)
              B.\(12cm\)
              C.\(11cm\)
              D.\(10cm\)
            • 5.

              如图,在\(□\)\(ABCD\)中,点\(E\)是边\(AD\)的中点,\(EC\)交对角线\(BD\)于点\(F\),则\(EF\)\(FC\)等于\((\)  \()\)


              A.\(3\):\(2\)     
              B.\(3\):\(1\)   
              C.\(1\):\(1\)     
              D.\(1\):\(2\)
            • 6.

              如图,在四边形\(ABCD\)中,\(∠B=90^{\circ}\),\(AC=4\),\(AB/\!/CD\),\(DH\)垂直平分\(AC\),点\(H\)为垂足,设\(AB=x\),\(AD=y\),则\(y\)关于\(x\)的函数关系用图象大致可以表示为(    )

              A.
              B.
              C.
              D.
            • 7.

              如图,在\(\triangle ABC\)中,\(D\),\(E\)分别在\(AB\),\(AC\)上,且\(D E/\!/BC\),则下列不成立的比例式是\((\)  \()\)


              A.\( \dfrac{AD}{DB}= \dfrac{AE}{CE}\)
              B.\( \dfrac{AD}{DB}= \dfrac{DE}{BC}\)
              C.\( \dfrac{AD}{AB}= \dfrac{AE}{AC}\)
              D.\( \dfrac{AB}{DB}= \dfrac{AC}{CE}\)
            • 8.
              如图,点\(F\)是矩形\(ABCD\)的边\(CD\)上一点,射线\(BF\)交\(AD\)的延长线于点\(E\),则下列结论错误的是\((\)  \()\)
              A.\( \dfrac {ED}{EA}= \dfrac {DF}{AB}\)
              B.\( \dfrac {DE}{BC}= \dfrac {EF}{FB}\)
              C.\( \dfrac {BC}{DE}= \dfrac {BF}{BE}\)
              D.\( \dfrac {BF}{BE}= \dfrac {BC}{AE}\)
            • 9.

              如图,在\({\triangle }ABC\)中,点\(D{,}E\)分别是\({AB}{,}{AC}\)边上的两个点,且\(\dfrac{{AE}}{{AB}}{=}\dfrac{{AD}}{{AC}}{=}\dfrac{1}{2}\),则\(S_{{\triangle }ADE}\):\(S_{{\triangle }ABC}\)的值为\(({  })\)


              A.\(1\):\(\sqrt{3}\)                      
              B.\(1\):\(2\)               
              C.\(1\):\(3\)               
              D.\(1\):\(4\)
            • 10.

              如图,已知在\(\triangle ABC\)中,\(\cos A=\dfrac{1}{3}\),\(BE\)、\(CF\)分别是\(AC\)、\(AB\)边上的高,联结\(EF\),那么\(\triangle AEF\)和\(\triangle ABC\)的周长比为\((\)      \()\)

              A.\(1︰3\);                  
              B.\(1︰2\);

              C.\(1︰4\);                
              D.\(1︰9\).
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