优优班--学霸训练营 > 知识点挑题
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            • 1. 如图,四边形\(ABCD\)内接于\(⊙O\),\(AD\)、\(BC\)的延长线相交于点\(E\),\(AB\)、\(DC\)的延长线相交于点\(F.\)若\(∠E+∠F=80^{\circ}\),则\(∠A=\) ______ \({\,\!}^{\circ}.\)
            • 2.

              如图,\(⊙O\)的内接四边形\(ABCD\)中,\(∠A=105^{\circ}\),则\(∠BOD\)等于                 

            • 3.

              如图,四边形 \(ABCD\)内接于\(⊙\)\(O\),四边形 \(ABCO\)是平行四边形,则\(∠\)\(ADC\)\(=(\)      \()\)

              A.\(45^{\circ}\)          
              B.\(50^{\circ}\)        
              C.\(60^{\circ}\)                
              D.\(75^{\circ}\)
            • 4.
              如图,四边形\(ABCD\)内接于半圆\(O\),已知\(∠ADC=140^{\circ}\),则\(∠AOC\)的大小是\((\)  \()\)
              A.\(40^{\circ}\)
              B.\(60^{\circ}\)
              C.\(70^{\circ}\)
              D.\(80^{\circ}\)
            • 5.
              如图,\(AB\)为\(⊙O\)的直径,弦\(CD⊥AB\)于\(H\),\(E\)为\(AB\)延长线上一点,\(CE\)交\(⊙O\)于点\(F\)
              \((1)\)求证:\(BF\)平分\(∠DFE\);
              \((2)\)若\(EF=DF\),\(BE=5\),\(AH= \dfrac {9}{4}\),求\(⊙O\)的半径.
            • 6.
              如图,四边形\(ABCD\)是\(⊙O\)的内接四边形,若\(∠C=65^{\circ}\),则\(∠A=\) ______ \({\,\!}^{\circ}.\)
            • 7.
              如图,四边形\(ABCD\)为\(⊙O\)的内接四边形,若\(∠BCD=110^{\circ}\),则\(∠BAD\)为\((\)  \()\)
              A.\(140^{\circ}\)
              B.\(110^{\circ}\)
              C.\(90^{\circ}\)
              D.\(70^{\circ}\)
            • 8.

              如图,\(⊙O\)的半径为\(3\),四边形\(ABCD\)内接于\(⊙O\),若\(2∠BAD=∠BCD\),则\(\overline {BD} \)的长为\((\)     \()\)


              A.\(π\)               
              B.\( \dfrac{3}{2}π \)       
              C.\(2π\)           
              D.\(3π\)
            • 9.

              如图,\(A\)\(B\)\(C\)\(D\)为\(⊙\)\(O\)上四点,若\(∠\)\(BOD\)\(=110º\),则\(∠\)\(A\)的度数是(    )  


              A.\(110º\)
              B.\(115º\)
              C.\(120º\)
              D.\(125º\)
            • 10.
              如图,在圆内接四边形\(ABCD\)中,\(∠A\)、\(∠C\)的度数之比为\(1\):\(2\),则\(∠A\)的度数为\((\)  \()\)
              A.\(30^{\circ}\)
              B.\(60^{\circ}\)
              C.\(70^{\circ}\)
              D.\(90^{\circ}\)
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