《2025年高效精练九年级数学上册苏科版》

10. (2024·济宁)如图,分别延长圆内接四边形ABCD的两组对边,延长线相交于点E,F.若$∠E=54^{\circ }41',∠F=43^{\circ }19'$,则$∠A$的度数为 (C)

A. $42^{\circ }$
B. $41^{\circ }20'$
C. $41^{\circ }$
D. $40^{\circ }20'$

答案： C

11. (2023·呼和浩特)如图,$△ABC$内接于$\odot O$且$∠ACB=90^{\circ }$,弦CD平分$∠ACB$,连接AD,BD.若$AB=5,AC=4$,则$BD=$$\frac{5\sqrt{2}}{2}$,$CD=$$\frac{7\sqrt{2}}{2}$.

答案： $\frac{5\sqrt{2}}{2}$ $\frac{7\sqrt{2}}{2}$

12. 现有直径为2的半圆O和一块等腰直角三角板.
(1) 将三角板如图1放置,锐角顶点P在圆上,斜边经过点B,一条直角边交圆于点Q,则BQ的长为$\sqrt{2}$;
(2) 将三角板如图2放置,锐角顶点P在圆上,斜边经过点B,一条直角边的延长线交圆于Q,则BQ的长为$\sqrt{2}$.

答案：
(1) $\sqrt{2}$
(2) $\sqrt{2}$

13. 如图,四边形ABCD内接于$\odot O$,分别延长BC,AD交于点E,$AB=8$,且$DC=DE$.
(1) 求证:$∠A=∠AEB$.
证明:$\because$四边形$ABCD$内接于$\odot O$,$\therefore ∠A + ∠BCD = 180^{\circ}$. $\because ∠BCD + ∠DCE = 180^{\circ}$,$\therefore ∠A = ∠DCE$,$\because DC = DE$,$\therefore ∠E = ∠DCE$,$\therefore ∠A = ∠AEB$;
(2) 若$∠EDC=90^{\circ }$,点C为BE的中点,求$\odot O$的半径.
解:如图,连接$AC$,$\because ∠EDC = 90^{\circ}$,$\therefore AC$是$\odot O$的直径,$\therefore ∠ABC = 90^{\circ}$,$\because ∠A = ∠AEB$,$\therefore AB = BE$. $\because AB = 8$,$\therefore BE = 8$. $\because$点$C$为$BE$的中点,$\therefore BC = \frac{1}{2}BE = 4$,在$Rt△ABC$中,$AC = \sqrt{AB^{2} + BC^{2}} = \sqrt{8^{2} + 4^{2}} = 4\sqrt{5}$,$\therefore \odot O$的半径为$2\sqrt{5}$.

答案：
(1) $\because$四边形$ABCD$内接于$\odot O$,$\therefore ∠A + ∠BCD = 180^{\circ}$. $\because ∠BCD + ∠DCE = 180^{\circ}$,$\therefore ∠A = ∠DCE$,$\because DC = DE$,$\therefore ∠E = ∠DCE$,$\therefore ∠A = ∠AEB$;
(2) 如图,连接$AC$,$\because ∠EDC = 90^{\circ}$,$\therefore AC$是$\odot O$的直径,$\therefore ∠ABC = 90^{\circ}$,$\because ∠A = ∠AEB$,$\therefore AB = BE$. $\because AB = 8$,$\therefore BE = 8$. $\because$点$C$为$BE$的中点,$\therefore BC = \frac{1}{2}BE = 4$,在$Rt△ABC$中,$AC = \sqrt{AB^{2} + BC^{2}} = \sqrt{8^{2} + 4^{2}} = 4\sqrt{5}$,$\therefore \odot O$的半径为$2\sqrt{5}$.

14. 已知四边形ABCD内一点E,若$EA=EB=EC=ED,∠BAD=70^{\circ }$,则$∠BCD$的度数为______$110^{\circ}$.

答案： $110^{\circ}$

15. (2023·北京)如图,圆内接四边形ABCD的对角线AC,BD交于点E,BD平分$∠ABC,∠BAC=∠ADB$.
(1) 求证:DB平分$∠ADC$,并求$∠BAD$的大小;90°
(2) 过点C作$CF// AD$交AB的延长线于点F,若$AC=AD,BF=2$,求此圆半径的长.4

答案：
(1) $\because ∠BAC = ∠ADB$,$∠BAC = ∠CDB$,$\therefore ∠ADB = ∠CDB$,$\therefore BD$平分$∠ADC$. $\because BD$平分$∠ABC$,$\therefore ∠ABD = ∠CBD$. $\because$四边形$ABCD$是圆内接四边形,$\therefore ∠ABC + ∠ADC = 180^{\circ}$,$\therefore ∠ABD + ∠CBD + ∠ADB + ∠CDB = 180^{\circ}$,$\therefore 2(∠ABD + ∠ADB) = 180^{\circ}$,$\therefore ∠ABD + ∠ADB = 90^{\circ}$,$\therefore ∠BAD = 180^{\circ} - 90^{\circ} = 90^{\circ}$;
(2) $\because ∠BAE + ∠DAE = 90^{\circ}$,$∠BAE = ∠ADE$,$\therefore ∠ADE + ∠DAE = 90^{\circ}$,$\therefore ∠AED = 90^{\circ}$. $\because ∠BAD = 90^{\circ}$,$\therefore BD$是圆的直径,$\therefore BD$垂直平分$AC$,$\therefore AD = CD$. $\because AC = AD$,$\therefore △ACD$是等边三角形,$\therefore ∠ADC = 60^{\circ}$. $\because BD ⊥ AC$,$\therefore ∠BDC = \frac{1}{2}∠ADC = 30^{\circ}$. $\because CF // AD$,$\therefore ∠F + ∠BAD = 180^{\circ}$,$\therefore ∠F = 90^{\circ}$. $\because$四边形$ABCD$是圆内接四边形,$\therefore ∠ADC + ∠ABC = 180^{\circ}$. $\because ∠FBC + ∠ABC = 180^{\circ}$,$\therefore ∠FBC = ∠ADC = 60^{\circ}$,$\therefore BC = 2BF = 4$. $\because ∠BCD = 90^{\circ}$,$∠BDC = 30^{\circ}$,$\therefore BC = \frac{1}{2}BD$. $\because BD$是圆的直径,$\therefore$圆的半径长是4.