《2025年学霸提优大试卷九年级数学全一册苏科版》

20. (8分)(盐城中考)如图,在$△ABC$中,$∠C= 90^{\circ },tanA= \frac {\sqrt {3}}{3},∠ABC$的平分线BD交AC于点D,$CD= \sqrt {3}$,求AB的长.
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答案： 在Rt△ABC中,∠C = 90°,tanA = $\frac{\sqrt{3}}{3}$,
∴∠A = 30°,
∴∠ABC = 60°.
∵BD是∠ABC的平分线,
∴∠CBD = ∠ABD = 30°.又
∵CD = $\sqrt{3}$,
∴BC = $\frac{CD}{tan30°}$ = 3.在Rt△ABC中,∠C = 90°,∠A = 30°,
∴AB = $\frac{BC}{sin30°}$ = 6.

21. (8分)(2024·长沙中考)如图,在$□ ABCD$中,对角线AC、BD相交于点O,$∠ABC= 90^{\circ }$.
(1)求证:$AC= BD$;
(2)点E在BC边上,满足$∠CEO= ∠COE$.若$AB= 6,BC= 8$,求CE的长及$tan∠CEO$的值.
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答案：
(1)因为四边形ABCD是平行四边形,且∠ABC = 90°,所以四边形ABCD是矩形,所以AC = BD.
(2)在Rt△ABC中,AB = 6,BC = 8,所以AC = $\sqrt{AB^{2}+BC^{2}}$ = $\sqrt{6^{2}+8^{2}}$ = 10.因为四边形ABCD是矩形,所以CO = $\frac{1}{2}$AC = 5,OB = OC.因为∠CEO = ∠COE,所以CE = CO = 5.过点O作OF⊥BC于点F,则CF = $\frac{1}{2}$BC = 4,所以EF = CE - CF = 5 - 4 = 1.在Rt△COF中,OF = $\sqrt{OC^{2}-CF^{2}}$ = $\sqrt{5^{2}-4^{2}}$ = 3,所以tan∠CEO = $\frac{OF}{EF}$ = 3.

22. (10分)(2024·西藏中考)如图,AB是$\odot O$的直径,C、D是$\odot O$上两点,连接AC、BC,CO平分$∠ACD,CE⊥DB$,交DB延长线于点E.
(1)求证:CE是$\odot O$的切线;
(2)若$\odot O$的半径为5,$sinD= \frac {3}{5}$,求BD的长.
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答案：
(1)
∵CO平分∠ACD,
∴∠ACO = ∠DCO = $\frac{1}{2}$∠ACD.
∵$\overset{\frown}{AD}$ = $\overset{\frown}{AD}$,
∴∠ABD = ∠ACD = 2∠ACO.
∵AO = CO,
∴∠ACO = ∠CAO,
∴∠COB = ∠ACO + ∠A = 2∠ACO,
∴∠ABD = ∠COB,
∴CO//DE.
∵CE⊥DE,
∴∠CED = 90°,
∴∠OCE = 180° - ∠CED = 90°,
∴OC⊥CE.
∵OC为⊙O的一条半径，
∴CE是⊙O的切线.
(2)
∵⊙O的半径为5,
∴AB = 2×5 = 10.
∵$\overset{\frown}{BC}$ = $\overset{\frown}{BC}$,
∴∠A = ∠D,
∴sinA = sinD = $\frac{3}{5}$.
∵AB为⊙O的直径，
∴∠ACB = 90°,
∴BC = AB×sinA = 10×$\frac{3}{5}$ = 6.
∵∠ECB + ∠BCO = ∠BCO + ∠ACO = 90°,
∴∠ECB = ∠ACO.
∵∠ACO = ∠A,
∴∠ECB = ∠A,
∴sin∠ECB = sinA = $\frac{3}{5}$,即$\frac{BE}{BC}$ = $\frac{3}{5}$,
∴BE = $\frac{3}{5}$×6 = $\frac{18}{5}$,
∴CE = $\sqrt{BC^{2}-BE^{2}}$ = $\sqrt{6^{2}-(\frac{18}{5})^{2}}$ = $\frac{24}{5}$.
∵sinD = $\frac{CE}{CD}$ = $\frac{3}{5}$,
∴CD = $\frac{5}{3}$×CE = $\frac{5}{3}$×$\frac{24}{5}$ = $\mathbf{8}$,
∴DE = $\sqrt{CD^{2}-CE^{2}}$ = $\sqrt{8^{2}-(\frac{24}{5})^{2}}$ = $\frac{32}{5}$,
∴BD = DE - BE = $\frac{32}{5}-\frac{18}{5}$ = $\frac{14}{5}$.