📚 IGCSE Maths: Circle Theorems – Chords, Angles and Tangents | IGCSE 数学：圆定理——弦、角与切线
Circle theorems are a core part of the IGCSE Maths geometry syllabus. They describe the relationships between angles, chords, tangents and radii in a circle. Mastering these rules allows you to solve angle problems without using a protractor, and they appear frequently in both Foundation and Higher tier papers. Circle theorem questions are usually worth 3 to 5 marks, and examiners award method marks for correct reasoning even when the final answer is wrong, so naming the theorem you used matters as much as the number itself.
圆定理是 IGCSE 数学几何部分的核心内容。它们描述了圆中角、弦、切线和半径之间的关系。掌握这些规则后，你可以不使用量角器就解决角度问题，这些定理在基础卷和进阶卷中都频繁出现。圆定理题目通常占 3 到 5 分，即使最终答案算错，只要推理过程正确也能拿到方法分，所以写出所用定理的名称，和算出数字同样重要。
1. Circle Terminology and Notation | 圆的术语与符号
Before using circle theorems, it is essential to learn the names of the key parts of a circle. The centre O is the fixed point from which all radii are drawn. A radius r is any line segment from the centre to a point on the circumference. A chord is a line segment joining two points on the circle. A diameter is a chord that passes through the centre, so its length is 2r. A tangent is a straight line that touches the circle at exactly one point. A secant is a line that cuts the circle at two points. An arc is part of the circumference, and a sector is the region between two radii and an arc. A segment is the region between a chord and an arc.
在使用圆定理之前，必须先学习圆各部分的名称。圆心 O 是所有半径的固定起点。半径 r 是从圆心到圆周上任意一点的线段。弦是连接圆上两点的线段。直径是经过圆心的弦，所以长度为 2r。切线是只与圆恰好相交于一点的直线。割线是与圆相交于两点的直线。弧是圆周的一部分，扇形是两条半径与一条弧之间的区域。弓形是弦与弧之间的区域。
In diagrams, angles such as ∠ABC mean the angle at vertex B, formed by lines AB and BC. The symbol ≅ means congruent lengths or angles, while ⊥ means perpendicular and ∥ means parallel. Equal-length chords are often marked with matching dashes, and equal angles with matching arcs.
在图形中，像 ∠ABC 这样的角表示顶点在 B、由线段 AB 与 BC 形成的角。符号 ≅ 表示长度或角度全等，⊥ 表示垂直，∥ 表示平行。等长的弦通常用相同数量的短划线标记，相等的角则用相同数量的弧线标记。
- Chord–arc vocabulary: the major arc is the longer arc, the minor arc is the shorter one; the major segment is bounded by the major arc.
- 弦与弧的词汇：较长的弧叫劣弧，较短的弧叫优弧；由优弧围成的区域叫优弓形。
- Why it matters: most lost marks come from misreading which arc an angle stands on, not from the theorem itself.
- 为什么重要：大多数丢分来自看错某个角所对的弧，而不是定理本身用错。
2. The Chord and Perpendicular Bisector Theorem | 弦与垂直平分线定理
dicular drawn from the centre of a circle to a chord bisects that chord. In symbols, if OM ⊥ e: the perpendicular bisector of a chord passes through the centre, which is why you can find the centre of any circle by drawing two chords and their perpendicular bisectors and seeing where the lines cross.
从圆心向弦作垂线，这条垂线会平分该弦。用符号表示：若 O 为圆心、M 在弦 AB 上且 OM ⊥ AB，则 AM = MB。逆命题同样成立：弦的垂直平分线必过圆心，因此只要画出两条弦及其垂直平分线，交点就是圆心。
Equal chords stand at equal distances from the centre, and chords at equal distances from the centre are equal. This gives a powerful link between lengths and angles that you can use inside a right-angled triangle formed by the radius, half the chord and the perpendicular distance.
等长的弦到圆心的距离相等；到圆心距离相等的弦长度相等。由此可以在由半径、半弦长和垂距构成的直角三角形中建立长度与角度之间的联系，这是解题的常用桥梁。
r² = d² + (½AB)²
This relationship lets you calculate a chord length when you know the radius and the distance from the centre, or find the radius when you know both the chord and the distance. It is a standard 3-mark application question in IGCSE papers.
已知半径和圆心到弦的距离时，这个关系式可以算出弦长；已知弦长和距离时，也能反过来求半径。在 IGCSE 试卷中，这是标准的 3 分应用题。
3. Angles at the Centre and at the Circumference | 圆心角与圆周角
g part of the circumference. If O is the centre and A, B, P are points on the circle, then ∠AOB = 2 × ∠APB, provided that O and P lie on the same side of chord AB.
同一段弧在圆心处所对的角，是它在圆周上任意一点所对角的 2 倍。若 O 为圆心，A、B、P 是圆上三点，且 O 与 P 位于弦 AB 的同侧，则 ∠AOB = 2 × ∠APB。
∠AOB = 2 × ∠APB
A direct consequence is that all angles subtended by the same arc in the same segment are equal. Two students sitting anywhere on the same major arc looking at the same chord will measure exactly the same angle.
由此可得出一个直接推论：同一弓形中，同一段弧所对的圆周角全都相等。无论站在同一优弧上的哪一点去看同一条弦，量出的角都完全相同。
- Direction of use: centre angle is bigger, so multiply the circumference angle by 2, not divide.
- 使用方向：圆心角更大，所以是把圆周角乘 2，而不是除 2。
- Watch the side: are on opposite sides of the chord, the angle at the centre is the reflex angle 360° − 2 × ∠APB.
- 注意位置：若圆心与圆周上的点在弦的两侧，此时圆心角是优角 360° − 2 × ∠APB。
4. The Angle in a Semicircle | 半圆上的圆周角
icircle is a right angle. If AB is a diameter and P is any other point on the circle, then ∠ This is simply the centre-angle theorem applied to the straight angle at the centre, since ∠AOB = 180° when AOB is a diameter.
半圆上的圆周角是直角。若 AB 是直径，P 是圆上任意其他一点，则 ∠APB = 90°。这其实就是圆心角定理的特殊情形：当 AOB 是一条直径时，圆心角 ∠AOB = 180°。
The converse is very useful in exam questions: if a triangle inscribed in a circle has a right angle, then the hypotenuse must be a diameter, so the midpoint of the hypotenuse is the centre of the circle. If you are told that a circle passes through the three vertices of a right-angled triangle, you can immediately state where its centre lies.
逆命题在考试中非常有用：若圆内接三角形有一个直角，则其斜边必为直径，因此斜边的中点就是圆心。若题目说一个圆经过直角三角形的三个顶点，你就可以直接指出圆心在何处。
angles in the same segment are equal. Two points P and Q on the same side of chord AB give ∠APB = ∠ this on the diagram before you calculate usually reveals which theorem the examiner wants.
相关的结论是：同一弓形中的圆周角相等。弦 AB 同侧的两点 P 和 Q 满足 ∠APB = ∠AQB。在计算之前先把这一点标在图上，通常就能看出命题人想考哪条定理。
5. Cyclic Quadrilaterals | 圆内接四边形
vertices all lie on a circle. Its opposite angles add up to 180°. If ABCD is cyclic, then ∠A + ∠C = 180° and ∠B + ∠D = 180°. These are called supplementary or co-interior-in-a-circle angles.
圆内接四边形是指四个顶点都在同一个圆上的四边形。它的对角之和为 180°。若 ABCD 为圆内接四边形，则 ∠A + ∠C = 180°，∠B + ∠D = 180°。这个性质称为对角互补。
∠A + ∠C = ∠B + ∠D = 180°
teral equals the interior opposite angle. If side AB is extended past B to a point E, then ∠CBE = ∠ t route when the question gives you an exterior angle rather than a fourth interior angle.
圆内接四边形的外角等于它的内对角。若把边 AB 延长到 B 外的点 E，则 ∠CBE = ∠ADC。当题目给出外角而不是第四个内角时，用这条性质通常最快。
- Proof check: if a question says "prove that ABCD is cyclic", show that one pair of opposite angles sums to 180°.
- 证明思路：若题目要求"证明 ABCD 四点共圆"，只需证明其中一对对角之和为 180°。
- Common trap: adjacent angles are not supplementary; only opposite pairs are.
- 常见陷阱：相邻的两个角并不互补，只有对角之和才为 180°。
6. Tangents and Radii | 切线与半径
A tangent to a circle is perpendicular to the radius at the point of contact. If the line PT touches the circle at T and O is the centre, then OT ⊥ PT. This single fact converts almost every tangent problem into a right-angled triangle problem, allowing you to use Pythagoras' theorem.
圆的切线在切点处与半径垂直。若直线 PT 与圆相切于 T，O 为圆心，则 OT ⊥ PT。仅凭这一条，几乎所有切线问题都能转化为直角三角形问题，从而使用勾股定理。
OT ⊥ PT and OP² = OT² + PT²
ntre bisects the angle between them. So if PA and PB are tangents from P, then PA = PB and ∠APO = ∠BPO. The two right-angled triangles OAP and OBP are congruent.
从同一点引出的两条切线长度相等，且该点与圆心的连线平分两条切线之间的夹角。因此若 PA、PB 是自 P 引出的切线，则 PA = PB，∠APO = ∠BPO，两个直角三角形 OAP 与 OBP 全等。
ternate segment, that is, the angle subtended by that chord at any point on the opposite arc. In symbols, with tangent TX and chord TA, ∠XTA = ∠TBA where B lies on the arc on the other side of TA.
第三条切线定理涉及切点处的弦与切线。切线与弦的夹角等于另一个弓形中的圆周角，也就是这条弦在对面弧上任意一点所对的角。用符号表示：切线 TX、弦 TA，B 位于 TA 另一侧的弧上，则 ∠XTA = ∠TBA。
7. Intersecting Chords and Secants (Extension) | 相交弦与割线定理（拓展）
chord are equal. If chords AB and CD meet at P inside the circle, then PA × PB = PC × PD.
当两条弦在圆内相交时，每条弦被交点分成的两段的乘积相等。若弦 AB 与 CD 在圆内交于点 P，则 PA × PB = PC × PD。
PA × PB = PC × PD
a-point results are not required by every IGCSE specification, so always check your own syllabus before relying on them, but they appear in some Extended and Additional Mathematics papers.
对于圆外一点，切线长度的平方等于整条割线与圆外部分之积：PT² = PA × PB。这些"点幂"结论并非所有 IGCSE 考纲都要求，使用前务必核对自己所属的考纲，但它们在部分进阶卷和附加数学试卷中会出现。
- Length rule of thumb: multiply the two segments on each line, then set the products equal.
- 长度口诀：把每条直线上的两段相乘，再令乘积相等。
- Unit care: all lengths must be in the same unit before you form the equation.
- 单位注意：列方程之前，所有长度必须先化成同一单位。
8. Proofs, Reasons and Exam Wording | 证明、理由与卷面表述
Nearly every circle theorem question requires a reason for each step. In IGCSE, the accepted phrases are short and standardised, so learn them exactly rather than describing the geometry in your own words.
几乎每道圆定理题都要求为每一步给出理由。在 IGCSE 中，可接受的表述简短且标准化，因此要逐字记住，而不是用自己的话描述图形关系。
| English reason | 中文对应 | When to use |
| Angle in a semicircle is 90° | 半圆上的圆周角是直角 | A diameter is given |
| Angle at centre is twice angle at circumference | 圆心角是圆周角的两倍 | Both angles stand on the same arc |
| Angles in the same segment are equal | 同弓形（同弧）圆周角相等 | Two points on the same arc |
| Opposite angles of a cyclic quadrilateral sum to 180° | 圆内接四边形对角互补 | Four points on a circle |
| Tangent is perpendicular to radius | 切线与半径垂直 | A tangent and a centre are given |
| Angle between tangent and chord equals angle in alternate segment | 弦切角等于另一弓形中的圆周角 | Tangent meets a chord at contact point |
lculation, and always give the theorem name before or after the number. A bare number like "102°" earns nothing; "∠ADC = 180° − 78° = 102° (opposite angles of a cyclic quadrilateral)" earns full marks.
把理由写在计算式旁边，并始终在数字前后给出定理名称。只写一个"102°"是拿不到分的；写成"∠ADC = 180° − 78° = 102°（圆内接四边形对角互补）"才能得满分。
9. Worked Exam-Style Examples | 真题风格例题精讲
Example 1: A, B, C and D lie on a circle in that order. ∠ABC = 78°. Find ∠ADC. Since ABCD is a cyclic quadrilateral, the opposite angles are supplementary, so ∠ADC = 180° − 78° = 102°.
例题一：A、B、C、D 依次位于同一个圆上，∠ABC = 78°，求 ∠ADC。因为 ABCD 是圆内接四边形，对角互补，所以 ∠ADC = 180° − 78° = 102°。
the centre of a circle, A and B lie on the circumference, and TA is a tangent at A. Given ∠TAB = 62°, find ∠AOB. First, ∠OAT = 90° because the tangent is perpendicular to the radius. Hence ∠OAB = 90° − 62° = 28°. Triangle OAB is isosceles since OA = OB = r, so ∠OBA = 28°. Then ∠AOB = 180° − 28° − 28° = 124°.
例题二：O 为圆心，A、B 在圆周上，TA 是过 A 的切线。已知 ∠TAB = 62°，求 ∠AOB。首先，因为切线与半径垂直，∠OAT = 90°；于是 ∠OAB = 90° − 62° = 28°。由于 OA = OB = r，三角形 OAB 是等腰三角形，所以 ∠OBA = 28°；因此 ∠AOB = 180° − 28° − 28° = 124°。
Check with the alternate segment theorem: ∠TAB = ∠ = 124°. Both methods agree, which is a good way to verify your answer under time pressure.
用弦切角定理验算：设 C 在另一段弧上，则 ∠TAB = ∠ACB = 62°，圆心角是它的 2 倍，即 2 × 62° = 124°，两种方法结果一致。在考试时间紧张时，这是很好的自查方式。
ving r = 10 cm. Always state the perpendicular-from-centre-bisects-the-chord reason first.
例题三：一条长 16 cm 的弦距圆心 6 cm，求半径。半弦长为 8 cm，故 r² = 6² + 8² = 36 + 64 = 100，得 r = 10 cm。作答时务必先写出"从圆心所作垂线平分弦"这一理由。
10. Common Mistakes and Exam Technique | 常见错误与应试技巧
The most frequent error is using the centre-angle theorem the wrong way round, dividing instead of doubling. A quick sanity check: the centre angle always looks wider in a correctly drawn diagram, so if your answer shrinks when it should grow, you have inverted the rule.
最常见的错误是把圆心角定理用反，把乘 2 写成除 2。一个快速的合理性检查：在画得准确的图中，圆心角看起来总是更宽，所以如果本该变大的结果反而变小了，就是公式用反了。
- Assume nothing is to scale: diagrams are illustrative, so never measure with a ruler or protractor.
- 不要假设图形按比例：示意图只作说明，绝不能用量角器或直尺去量。
- Mark what you know: copy the diagram and label every given angle, equal radius and equal tangent before starting.
- 先标注已知条件：动笔前先把图形抄下来，标出所有已知角、相等半径和相等切线。
- Show the equation: write "180° − x" or "2 × x" explicitly rather than only the final number.
- 写出算式：把"180° − x"或"2 × x"明确写出来，而不能只给最终数字。
- Check reasonableness: an angle in a triangle must be between 0° and 180°, and a cyclic quadrilateral angle must be under 180°.
- 检查合理性：三角形内角必须在 0° 到 180° 之间，圆内接四边形中的角也必须小于 180°。
- Answer the question asked: if it asks for a reason rather than a value, give the theorem name, not a number.
- 看清问题要求：若题目要求说明理由而非求值，就要写出定理名称，而不是给一个数字。
11. Quick Revision Summary | 快速复习总表
Circle theorem questions reward pattern recognition more than calculation. Once you have identified the configuration on the diagram, the route to the answer is usually two or three lines long.
圆定理题更看重对图形模式的识别，而不是计算量。一旦在图上认出属于哪一类结构，通往答案的路径通常只有两三行。
| Configuration | Rule | 中文要点 |
| Centre + two circumference points | ∠centre = 2 × ∠circumference | 圆心角是圆周角的两倍 |
| Diameter with a third point | Angle = 90° | 半圆上的角是直角 |
| Two points on the same arc | Angles in the same segment are equal | 同弓形的圆周角相等 |
| Four points on a circle | Opposite angles sum to 180° | 对角互补 |
| Tangent at a point | Tangent ⊥ radius | 切线与半径垂直 |
| Two tangents from one point | Equal lengths, equal angles at the centre line | 切线等长，圆心连线平分夹角 |
| Tangent meeting a chord | Alternate segment angles are equal | 弦切角等于另一弓形的圆周角 |
| Perpendicular from centre to chord | Bisects the chord; r² = d² + (½AB)² | 平分弦；r² = d² + 半弦长² |
If you can draw all eight configurations from memory and state the rule for each in one sentence, you are ready for any circle theorem question that IGCSE can throw at you. Practise past-paper questions under timed conditions, and always write the reason next to every calculated angle.
如果能在纸上凭记忆画出这八种图形，并用一句话说出各自的定理，那么应对 IGCSE 中任何圆定理题目都不成问题。建议限时刷历年真题，并且养成在每个算出的角旁边写明理由的习惯。
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