《2025年提分教练九年级数学上册人教版》

19. (12 分)已知 $AB$ 是$\odot O$ 的直径,弦 $CD$ 与 $AB$ 相交,$\angle BAC = 38^{\circ}$.
(1)如图①,若 $D$ 为$\overset{\frown}{AB}$ 的中点,求$\angle ABC$ 和$\angle ABD$ 的度数.
(2)如图②,过点 $D$ 作$\odot O$ 的切线,与 $AB$ 的延长线交于点 $P$,若 $DP // AC$,求$\angle OCD$ 的度数.

答案： 19.解:
(1)连接OD，如图①.
∵AB是$\odot O$的直径，$\therefore ∠ACB = 90^{\circ }$，$\because ∠BAC = 38^{\circ }$，$\therefore ∠ABC = 90^{\circ } - 38^{\circ } = 52^{\circ }$.
∵D为$\widehat {AB}$的中点，$∠AOB = 180^{\circ }$，$\therefore ∠AOD = 90^{\circ }$，$\therefore ∠ABD = 45^{\circ }$.
(2)连接OD，如图②.
∵DP切$\odot O$于点D，$\therefore OD⊥DP$，即$∠ODP = 90^{\circ }$，$\because DP// AC$，$\therefore ∠P = ∠BAC = 38^{\circ }$，
∵∠AOD是$△ODP$的一个外角，$\therefore ∠AOD = ∠P + ∠ODP = 128^{\circ }$，$\therefore ∠ACD = 64^{\circ }$，$\because OC = OA$，$\therefore ∠OCA = ∠BAC = 38^{\circ }$，$\therefore ∠OCD = ∠ACD - ∠OCA = 64^{\circ } - 38^{\circ } = 26^{\circ }$.

20. (12 分)如图,$AB$ 是$\odot O$ 的直径,$AC$ 为弦,$\angle BAC$ 的平分线交$\odot O$ 于点 $D$,过点 $D$ 的切线交 $AC$ 的延长线于点 $E$.
求证：(1)$DE \perp AE$；
(2)$AE + CE = AB$.

答案： 20.证明:
(1)连接OD，如图所示.$\because OA = OD$，AD平分$∠BAC$，$\therefore ∠OAD = ∠ODA$，$∠CAD = ∠OAD$，$\therefore ∠CAD = ∠ODA$，$\therefore AE// OD$.
∵DE是$\odot O$的切线，$\therefore ∠ODE = 90^{\circ }$，$\therefore ∠E = 90^{\circ }$，即$DE⊥AE$；
(2)过点D作$DM⊥AB$于点M，连接CD、DB，如图所示.
∵AD平分$∠BAC$，$DE⊥AE$，$DM⊥AB$，$\therefore DE = DM$.在$Rt△DAE$和$Rt△DAM$中，$\left\{\begin{array}{l} DE = DM,\\ AD = AD,\end{array}\right. $$\therefore Rt△DAE\cong Rt△DAM(HL)$，$\therefore AE = AM$.$\because ∠EAD = ∠MAD$，$\therefore \widehat {CD}=\widehat {BD}$，$\therefore CD = BD$.在$Rt△DEC$和$Rt△DMB$中，$\left\{\begin{array}{l} DE = DM,\\ CD = BD,\end{array}\right. $$\therefore Rt△DEC\cong Rt△DMB(HL)$，$\therefore CE = BM$，$\therefore AE + CE = AM + BM = AB$.

21. (13 分)如图,已知点 $A$,$B$,$C$ 在半径为 4 的$\odot O$ 上,过点 $C$ 作$\odot O$ 的切线交 $OA$ 的延长线于点 $D$.
(1)若$\angle ABC = 29^{\circ}$,求$\angle D$ 的度数.
(2)若$\angle D = 30^{\circ}$,$\angle BAO = 15^{\circ}$,作 $CE \perp AB$ 于点 $E$,求：
① $BE$ 的长；
② 四边形 $ABCD$ 的面积.

答案： 21.解:
(1)连接OC，
∵CD为$\odot O$的切线，$\therefore OC⊥CD$，$\therefore ∠OCD = 90^{\circ }$，又$∠AOC = 2∠ABC = 29^{\circ }×2 = 58^{\circ }$，$\therefore ∠D = 90^{\circ } - 58^{\circ } = 32^{\circ }$.
(2)①连接OB，在$Rt△OCD$中，$\because ∠D = 30^{\circ }$，$OC = 4$，$\therefore ∠DOC = 60^{\circ }$，$CD = 4\sqrt {3}$，$\because ∠BAO = 15^{\circ }$，$\therefore ∠OBA = 15^{\circ }$，$\therefore ∠AOB = 150^{\circ }$，$\therefore ∠BOC = 150^{\circ } - 60^{\circ } = 90^{\circ }$，$\therefore △BOC$为等腰直角三角形，根据勾股定理，得$BC = 4\sqrt {2}$，$\because ∠ABC=\frac{1}{2}∠AOC = 30^{\circ }$，$\therefore$在$Rt△CBE$中，$CE=\frac{1}{2}BC = 2\sqrt {2}$，根据勾股定理，得$BE = 2\sqrt {6}$.②作$BH⊥OA$交AO的延长线于点H，如图，$\because ∠BOH = 180^{\circ } - ∠AOB = 30^{\circ }$，$\therefore BH=\frac{1}{2}OB = 2$，
∴四边形ABCD的面积$=S_{\triangle OBC}+S_{\triangle OCD}-S_{\triangle OAB}=\frac{1}{2}×4×4+\frac{1}{2}×4×4\sqrt {3}-\frac{1}{2}×4×2 = 8\sqrt {3}+4$.