1[中]如图，量角器的直径与直角三角板ABC的斜边AB重合，其中量角器0刻度线的端点N与点A重合，射线CP从CA处出发，以每秒$3°$的速度顺时针旋转，CP与量角器的半圆弧交于点E，第12秒时，点E在量角器上对应的读数是（）

A.$18°$
B.$36°$
C.$72°$
D.$144°$

（第1题图）
（第2题图）

1.C 【解析】如图,连接 OE.
$\because$ 射线 CP 从 CA 处出发,以每秒 $3°$ 的速度顺时针旋转,
$\therefore$ 第 12 秒时, $∠ACE = 3° × 12 = 36°$. $\because ∠ACB = 90°$,$\therefore$ 点 C 在以 AB 为直径的圆上,即点 C 在 $\odot O$ 上,
$\therefore ∠EOA = 2∠ECA = 2×36° = 72°$,故选 C.

2[2026山东威海期末,中]足球射门,不考虑其他因素,仅考虑射门点到球门MN的张角大小时,张角越大,射门越好.如图的正方形网格中,点A,B,C,M,N均在格点上,球员带球沿AB方向进攻,最好的射门点是 ()

A.点A
B.点B或点C
C.线段AB(异于端点)上的点
D.线段BC(异于端点)上的点

2.D 【解析】设小正方形的边长为 1. 如图,连接 CN, BM, BN, AM, AN. 根据勾股定理可得 $CN = \sqrt{2^2 + 4^2} = 2\sqrt{5}$, $CB = \sqrt{1^2 + 1^2} = \sqrt{2}$, $BN = \sqrt{3^2 + 3^2} = 3\sqrt{2}$, $\therefore CB^2 + BN^2 = CN^2$, $\therefore ∠CBN = 90°$. $\because ∠CMN = 90°$, $\therefore M$, B 都在以 CN 为直径的圆上, $\therefore C, M, N, B$ 四点共圆, $\therefore ∠MCN = ∠MBN$,则可得点 A 和线段 AB (异于端点)上的点都在圆外, $\therefore$ 点 A 和线段 AB (异于端点)上的点到 MN 的张角都小于 $∠MCN$,线段 BC(异于端点)上的点到 MN 的张角都大于 $∠MCN$, $\therefore$ 最好的射门点是线段 BC (异于端点)上的点,故选 D.

3 构造同弧所对的圆周角 [2025 浙江杭州质检，中]如图，圆内接$△ ABC$的外角$∠ ACH$的平分线与圆相交于点$D$，$DP ⊥ AC$，垂足为$P$，$DH ⊥ BH$，垂足为$H$，有下列结论：①$\overset{\frown}{AD} = \overset{\frown}{BD}$；②$AP=BH$；③$\overset{\frown}{AB} = \overset{\frown}{BC}$，其中一定成立的结论有（）


A.1个
B.2个
C.3个
D.0个

3.B 【解析】连接 BD,如图.
$\because CD$ 平分 $∠ACH, DP ⊥ AC, DH ⊥ BH, \therefore DP = DH$.
$\because ∠DAC$ 和 $∠DBC$ 是 $\overset{\frown}{CD}$ 所对的圆周角, $\therefore ∠DAC = ∠DBC$,即 $∠DAP = ∠DBH$. 在 $△ ADP$ 和 $△ BDH$ 中,
$\begin{cases} ∠DAP = ∠DBH, \\ ∠APD = ∠BHD = 90°, \\ DP = DH, \end{cases}$ $\therefore △ ADP ≅ △ BDH$ (AAS), $\therefore AD = DB, AP = BH, \therefore \overset{\frown}{AD} = \overset{\frown}{BD}$,故①②一定成立. 当 $AB = BC$ 时, $\overset{\frown}{AB} = \overset{\frown}{BC}$,故③不一定成立. 综上所述,一定成立的结论有 2 个,故选 B.

4 连半径,构造圆心角 [2025 江苏淮安期中,中]
如图所示,AB为$\odot O$的直径,点C在$\odot O$上,且$OC ⊥ AB$,弦CD与线段OB相交于点E,满足$∠ AEC = 65°$,连接AD,则$∠ BAD =$ °.

(第4题图)
(第5题图)

4.20 【解析】连接 OD,如图.
$\because OC ⊥ AB, \therefore ∠COE = 90°$.
$\because ∠AEC = 65°, \therefore ∠OCE = 180° - 90° - 65° = 25°$. $\because OC = OD, \therefore ∠ODC = ∠OCE = 25°$,
$\therefore ∠DOC = 180° - 25° - 25° = 130°, \therefore ∠BOD = ∠DOC - ∠COE = 40°$,
$\therefore ∠BAD = \frac{1}{2}∠BOD = 20°$. 故答案为 20.

5 遇直径构造90度的圆周角 [2026北京海淀区质检,中]如图,AB是$\odot O$的直径,D在弦BC的延长线上,$CD=BC$,DA的延长线交$\odot O$于点E,若$∠ DAB=130°$,连接CE,则$∠ E$的度数为.

5.$25°$ 【解析】如图,连接 AC. $\because AB$ 是 $\odot O$ 的直径,$\therefore ∠ACB = 90°, \therefore AC ⊥ BD$. $\because CD = BC, \therefore AC$ 是 BD 的垂直平分线, $\therefore AD = AB$. $\because ∠DAB = 130°, \therefore ∠B = ∠D = \frac{1}{2}×(180° - 130°) = 25°$,
$\therefore ∠E = ∠B = 25°$,故答案为 $25°$.

6[2026江苏盐城期中,中]如图,矩形ABCD的四个顶点在$\odot O$上,点E在$\odot O$上,$\overset{\frown}{AE}=\overset{\frown}{AB}$,连接CE,AE,BE.
(1)求证:$BC=CE$;
(2)若AD分别交EB,EC于点F,G,$AF=1$,$FG=3$,求$\odot O$的半径长.

6.(1)【证明】$\because$ 矩形 ABCD 的四个顶点在 $\odot O$ 上,
$\therefore AB = CD, AD = BC, \therefore \overset{\frown}{AB} = \overset{\frown}{CD}$. $\because \overset{\frown}{AE} = \overset{\frown}{AB}$,
$\therefore \overset{\frown}{AE} = \overset{\frown}{CD}, \therefore \overset{\frown}{AE} + \overset{\frown}{DE} = \overset{\frown}{CD} + \overset{\frown}{DE}$,即 $\overset{\frown}{AD} = \overset{\frown}{CE}$,
$\therefore AD = CE, \therefore BC = CE$.
(2)【解】如图,连接 AC.
由(1)知 $\overset{\frown}{AE} = \overset{\frown}{CD}, \therefore AE = CD$. $\because ∠ADC = 90°, \therefore AC$ 是 $\odot O$ 的直径, $\therefore ∠AEC = 90°$. $\because BC = CE, \therefore ∠BEC = ∠CBE$. $\because AD // BC$,
$\therefore ∠EFG = ∠CBE, \therefore ∠EFG = ∠BEC, \therefore GE = GF = 3$, $\therefore$ 在 $\mathrm{Rt}△ AEG$ 中, $AE = \sqrt{AG^2 - EG^2} = \sqrt{(1+3)^2 - 3^2} = \sqrt{7}$, $\therefore CD = \sqrt{7}$. $\because ∠AGE = ∠CGD, ∠AEG = ∠CDG = 90°, AE = CD, \therefore △ AEG ≅ △ CGD$ (AAS), $\therefore GD = GE = 3, \therefore AD = 7, \therefore$ 在 $\mathrm{Rt}△ ACD$ 中, $AC = \sqrt{AD^2 + CD^2} = \sqrt{7^2 + (\sqrt{7})^2} = 2\sqrt{14}, \therefore \odot O$ 的半径为 $\sqrt{14}$.

7[2026安徽合肥三模,中]如图,AB是$\odot O$的直径,$CD=CB,CE ⊥ AB$于点$E$,连接$AC,BD$交$CE$于点$F$.
(1)求证:$CF=BF$.
(2)若$CD=4\sqrt{5},AC=8\sqrt{5}$,求弦$BD$的长.

7.(1)【证明】$\because AB$ 是 $\odot O$ 的直径, $\therefore ∠ACB = 90°$,
$\therefore ∠A = 90° - ∠ABC$. $\because CE ⊥ AB, \therefore ∠CEB = 90°$,
$\therefore ∠ECB = 90° - ∠ABC, \therefore ∠ECB = ∠A$. $\because CD = CB, \therefore \overset{\frown}{CD} = \overset{\frown}{CB}, \therefore ∠DBC = ∠A, \therefore ∠ECB = ∠DBC, \therefore CF = BF$.
(2)【解】连接 OC,交 BD 于点 G,如图.
$\because \overset{\frown}{BC} = \overset{\frown}{CD}, \therefore OC ⊥ BD, BD = 2BG$. $\because ∠ACB = 90°$,
$BC = CD = 4\sqrt{5}, AC = 8\sqrt{5}$,
$\therefore AB = \sqrt{AC^2 + BC^2} = \sqrt{(8\sqrt{5})^2 + (4\sqrt{5})^2} = 20$,
$\therefore \odot O$ 的半径为 10. 设 $OG = x$, 则 $CG = 10 - x$. 由勾股定理,得 $BG^2 = OB^2 - OG^2 = BC^2 - CG^2$,即 $10^2 - x^2 = (4\sqrt{5})^2 - (10 - x)^2$, 解得 $x = 6, \therefore BG = \sqrt{10^2 - 6^2} = 8, \therefore BD = 16$.