📚 Circle Theorems for IGCSE Mathematics | IGCSE 数学圆定理精讲
Circle theorems are geometric rules that describe the relationships between angles, chords, tangents and radii in a circle. In IGCSE Mathematics they are essential for solving multi-step angle problems and for justifying reasoning without measurement, and they appear in Paper 2 and Paper 4, often mixed with algebra and the angle properties of triangles and parallel lines.
圆定理是描述圆中角、弦、切线和半径之间关系的几何规则。在 IGCSE 数学中，这些定理是解决多步角度问题和在无需测量的情况下证明推理的关键，它们会出现在 Paper 2 和 Paper 4 中，通常与代数以及三角形和平行线的角度性质综合考查。
1. Circle Vocabulary You Need | 你需要掌握的圆的相关术语
Before using circle theorems, make sure you can identify the parts of a circle: centre, radius, chord, tangent, diameter, arc, segment and sector. Every theorem is named after one of these objects, so a shaky vocabulary leads directly to choosing the wrong rule.
在使用圆定理之前，请确保你能识别圆的各个部分：圆心、半径、弦、切线、直径、弧、弓形和扇形。每一个定理都以这些对象之一命名，所以术语不牢会直接导致选错定理。
In most IGCSE diagrams, the centre is labelled O, and points on the circumference are labelled with capital letters such as A, B, C and P. A tangent line usually carries a letter such as T, and the point where the tangent touches the circle is called the point of contact.
在大多数 IGCSE 图形中，圆心标记为 O，圆周上的点用大写字母表示，例如 A、B、C 和 P。切线通常用 T 这样的字母标注，切线与圆的接触点称为切点。
| Term 术语 | Meaning 含义 |
| Radius 半径 | Line segment from the centre to the circumference. 从圆心到圆周的线段。 |
| Chord 弦 | Line segment joining two points on the circle. 连接圆上两点的线段。 |
| Tangent 切线 | Line that touches the circle at exactly one point. 与圆只有一个交点的直线。 |
| Arc 弧 | Part of the circumference. 圆周的一部分。 |
| Diameter 直径 | A chord passing through the centre; twice the radius. 经过圆心的弦，长度是半径的两倍。 |
| Segment 弓形 | Region between a chord and an arc (minor or major). 弦与弧之间的区域（劣弓形或优弓形）。 |
| Sector 扇形 | Region bounded by two radii and an arc. 由两条半径和一段弧围成的区域。 |
| Subtend 对（角） | An angle subtends an arc or chord when its endpoints lie on that arc or chord. 当角的端点落在某弧或弦上时，称该角对这段弧或弦。 |
| Cyclic quadrilateral 圆内接四边形 | A quadrilateral whose four vertices lie on a circle. 四个顶点都在一个圆上的四边形。 |
Drawing and labelling a circle before solving a question helps you avoid confusing arcs and segments. Mark the centre O, mark any equal radii, and shade or highlight the arc that the question is really about.
在解题前画出并标记圆，可以帮助你避免混淆弧和弓形。标出圆心 O，标出所有相等的半径，并把你真正需要的那段弧高亮或标记出来。
2. Where Circle Theorems Appear in the Exam | 圆定理在考试中的出现位置
Circle theorems are examined in both the core and extended routes. Short, single-theorem questions appear in Paper 2, while longer multi-step problems that combine two or three theorems with algebra appear in Paper 4. The same content also appears in Edexcel IGCSE papers and in Cambridge IGCSE papers that share the 0580 syllabus content.
圆定理在核心和拓展两条路径中都会被考查。单一 theorem 的短题出现在 Paper 2，而把两三个定理与代数结合的多步长题出现在 Paper 4。同样的内容也会出现在 Edexcel IGCSE 试卷以及与之共享 0580 考纲内容的 Cambridge IGCSE 试卷中。
A typical question gives you a diagram with several angles marked and asks you to find one unknown angle, with the instruction "Give a reason for each stage of your working." Marks are usually split: one mark for the correct value and one mark for each correct reason, so a right answer with no reason can lose half the marks.
典型题目会给出一个标有几个角的图形，要求你求出一个未知角，并附上 "Give a reason for each stage of your working." 的指令。分数通常是拆分的：数值正确得 1 分，每一条正确理由各得 1 分，所以答案对但没有理由可能丢掉一半的分。
Circle theorems also mix with algebra. You may be told that an angle is (3x + 10)° and another is (5x − 30)°, then asked to form an equation from a circle theorem and solve it. The geometry chooses the equation; the algebra finishes the job.
圆定理也会与代数综合。题目可能告诉你一个角是 (3x + 10)°，另一个角是 (5x − 30)°，然后要求你利用某个圆定理列出方程并求解。几何决定方程的形式，代数负责收尾。
Finally, circle theorems connect to the angle properties of triangles and parallel lines: the angles in a triangle sum to 180°, base angles of an isosceles triangle are equal, vertically opposite angles are equal, and alternate and corresponding angles appear when a tangent runs parallel to a chord.
最后，圆定理与三角形和平行线的角度性质相连：三角形内角和为 180°，等腰三角形的底角相等，对顶角相等，而当切线与弦平行时，会出现内错角和同位角。
3. Theorem 1: The Angle in a Semicircle | 定理一：半圆上的圆周角是直角
diameter, then the angle at C is 90°. In symbols, angle ACB = 90°. The reason to write is "angle in a semicircle is 90°" or "angle subtended by a diameter is 90°".
如果 A、B、C 都在圆上，且 AB 是直径，那么 C 处的角就是 90°。用符号表示即 angle ACB = 90°。书写理由是 "angle in a semicircle is 90°" 或 "angle subtended by a diameter is 90°"。
angle OCA = a and angle OBC = angle OCB = b. In triangle ABC, the angles sum to 180°, so 2a + 2b = 180°, giving a + b = 90°, which is exactly angle ACB.
证明很短，值得掌握。设 O 为圆心。则 OA、OB、OC 都是半径，所以三角形 OAC 和 OBC 都是等腰三角形。记 angle OAC = angle OCA = a，angle OBC = angle OCB = b。在三角形 ABC 中，内角和为 180°，所以 2a + 2b = 180°，得 a + b = 90°，这正是 angle ACB。
In exam diagrams this theorem is often hidden. Look for a triangle whose longest side passes through the centre, or a question that gives you a diameter and an angle of 90° somewhere else, which is a strong hint that the converse is being used.
在考试图形中，这个定理常常被"藏起来"。注意寻找最长边经过圆心的三角形，或者题目给出直径并在别处给出一个 90° 角——这是一个很强的提示：题目在用它的逆定理。
ide of the right-angled triangle is 5 cm, the third side is 12 cm, because 5² + 12² = 13².
常见的变式是把它与勾股定理结合：若直径为 13 cm，直角三角形的一条直角边为 5 cm，则第三边为 12 cm，因为 5² + 12² = 13²。
4. Theorem 2: The Angle at the Centre | 定理二：圆心角是圆周角的两倍
The angle subtended by an arc at the centre is twice the angle subtended by the same arc at the circumference, provided both angles stand on the same arc. Write "angle at centre is twice angle at circumference".
同一段弧在圆心处所对的角，是它在圆周上所对的角的两倍，前提是两个角都对着同一段弧。书写理由是 "angle at centre is twice angle at circumference"。
angle AOB = 2 × angle ACB
A = x. The exterior angle at O of triangle OAC equals the sum of the two opposite interior angles, so angle AOD = 2x, where D is the point where AO is extended. Repeating this on the other side gives the full result.
证明用到等腰三角形和三角形的外角。因为 OA 与 OC 是半径，三角形 OAC 是等腰三角形，所以 angle OAC = angle OCA = x。三角形 OAC 在 O 处的外角等于两个不相邻内角之和，因此 angle AOD = 2x，其中 D 是 AO 延长后与圆的交点。在另一侧重复同样的推理，就得到完整结论。
Be careful with the reflex case: if the angle at the centre you are given is reflex, the relationship still holds, but the acute angle at the centre is 360° minus the reflex angle. Exam questions often rely on this, so always decide which of the two centre angles is being described.
注意优角的情形：如果题目给出的圆心角是优角，关系仍然成立，但劣角版的圆心角等于 360° 减去优角。考题常利用这一点，因此一定要判断清楚题目描述的是哪一个圆心角。
This theorem is also the parent of two others: the angle in a semicircle is the special case where the centre angle is 180°, and angles in the same segment are equal because they all share the same doubling relationship with the centre angle.
这个定理还是另外两个定理的"母定理"：半圆上的圆周角是圆心角为 180° 的特殊情形；同弧圆周角相等，则是因为它们对同一个圆心角都有"两倍"关系。
5. Theorem 3: Angles in the Same Segment | 定理三：同弧圆周角相等
le and C and D are on the same side of chord AB, then angle ACB = angle ADB. The reason is "angles in the same segment are equal".
在同一弓形中，同一段弧所对的圆周角相等。若 A、B、C、D 都在圆上，且 C、D 位于弦 AB 的同侧，则 angle ACB = angle ADB。理由是 "angles in the same segment are equal"。
The word "same" does three jobs here and all three matter. The angles must subtend the same chord, they must have their vertices on the same arc, and they must lie on the same side of that chord. If the two vertices are on opposite sides of the chord, the angles are supplementary, not equal.
这里的 "same" 有三层含义，每一层都很关键：两角必须对同一条弦；顶点必须落在同一段弧上；必须位于弦的同一侧。如果两个顶点在弦的两侧，那么这两个角是互补的，而不是相等。
This theorem is extremely useful when a diagram shows a "bow tie" or "butterfly" ch is why they are similar and why the intersecting chords result PA × PB = PC × PD holds.
当图形出现两条弦在圆内相交的"蝴蝶形"时，这个定理格外好用。由此形成的两个三角形在相同弓形中有相等的角，因此它们相似，这也正是相交弦结论 PA × PB = PC × PD 成立的原因。
Remember to name the correct endpoints. Writing "angle ACB = angle ADB" is exact; writing "the two angles are equal because they are in the same segment" t naming them is usually accepted, but naming them protects you if the diagram is crowded.
切记写出正确的端点名称。写成 "angle ACB = angle ADB" 最精确；只写 "这两个角相等，因为它们在同一弓形中" 通常也可接受，但在图形拥挤时，写出名称能保护你的分数。
6. Theorem 4: Cyclic Quadrilaterals | 定理四：圆内接四边形对角互补
If all four vertices of a quadrilateral lie on a circle, the quadrilateral is cyclic, and each pair of opposite angles sums to 180°. The reason is "opposite angles of a cyclic quadrilateral sum to 180°" or simply "cyclic quadrilateral".
如果四边形的四个顶点都在同一个圆上，它就是圆内接四边形，每一组对角之和都为 180°。理由是 "opposite angles of a cyclic quadrilateral sum to 180°"，或简写为 "cyclic quadrilateral"。
angle A + angle C = 180° and angle B + angle D = 180°
ubtended by chord CD... combining the two arcs gives 2 × (angle A + angle C) = 360°, so angle A + angle C = 180°.
证明是定理二的巧妙应用。设 O 为圆心。弦 BC 在圆心所对的角是 2 × angle A，各段弧对应的圆心角合起来得到 2 × (angle A + angle C) = 360°，因此 angle A + angle C = 180°。
The corollary is equally examinable: an exterior angle of a cyclic quadrilateral equals the interior opposite angle. If a side of the quadrilateral is extended, the angle between the extended side and the adjacent side equals the angle at the opposite vertex.
它的推论同样常考：圆内接四边形的一个外角等于与它不相邻的内对角。如果把四边形的一条边延长，延长线与相邻边所成的角，等于对面顶点的角。
A special case is the rectangle, which is always cyclic with all four angles equal to 90°. A parallelogram is only cyclic when it is a rectangle, because opposite angles must both be 90° for them to sum to 180°.
一个特例是矩形：它总是圆内接四边形，四个角都是 90°。平行四边形只有在是矩形时才是圆内接四边形，因为对角必须都等于 90° 才能相加为 180°。
7. Theorem 5: The Tangent and the Radius | 定理五：切线与半径垂直
A tangent to a circle is perpendicular to the radius drawn to the point of contact. Write "tangent is perpendicular to radius (at the point of contact)". This gives you a right angle, which you can then feed into Pythagoras or into angle sums in a triangle.
圆的切线与过切点的半径垂直。书写理由是 "tangent is perpendicular to radius (at the point of contact)"。这为你提供了一个直角，之后可以代入勾股定理或在三角形中做角度求和。
The converse also works both ways in exam geometry: if a line is perpendicular to a radius at a point on the circumference, then that line is a tangent. And if a line is a tangent, then the distance from the centre to the line equals the radius, which is the key idea in coordinate-geometry versions of the question.
逆定理在考试几何中双向可用：如果一条直线在圆周某点处与半径垂直，那么这条直线就是切线；如果一条直线是切线，那么圆心到它的距离等于半径——这是坐标几何版本题目的核心思路。
Typical applications include: finding the length of a tangent from an external point using Pythagoras, finding an unknown angle in a triangle formed by two tangents and a chord, and proving that two lines are parallel by showing both are perpendicular to the same radius.
典型应用包括：用勾股定理求圆外一点到切点的切线长；求由两条切线与一条弦构成的三角形中的未知角；以及通过证明两条直线都与同一条半径垂直来证明它们平行。
One exam trap: the diagram may show a line that looks as though it merely touches the circle but no tangent statement is given. Never assume tangency unless the question says so or you can justify it.
一个考试陷阱：图形中可能出现一条看起来"刚好碰到"圆的线，但题目并未说它是切线。除非题目说明或你能给出证明，否则绝不要假设相切。
8. Theorem 6: Two Tangents from an External Point | 定理六：外一点引两条切线
t P to a circle with centre O, touching the circle at A and B, then PA = PB. The reason is "tangents from an external point are equal in length".
从圆外一点 P 向圆心为 O 的圆引两条切线，切点分别为 A 和 B，则 PA = PB。理由是 "tangents from an external point are equal in length"。
nd, OP also bisects the angle between the two radii OA and OB, so angle AOP = angle POB. Both follow because triangles OAP and OBP are congruent: OA = OB (radii), PA = PB (tangents), and angle OAP = angle OBP = 90° (tangent meets radius).
同一图形还附带两个结论。第一，直线 OP 平分两条切线之间的夹角，即 angle APO = angle OPB。第二，OP 也平分两条半径 OA 与 OB 之间的夹角，即 angle AOP = angle POB。两者都源于三角形 OAP 与 OBP 全等：OA = OB（半径），PA = PB（切线），且 angle OAP = angle OBP = 90°（切线垂直于半径）。
The shape OAPB is a kite with two right angles and a line of symmetry along OP. Recognising this kite saves time: any question that gives you two tangents from one point can be solved with symmetry and Pythagoras rather than with a long chain of angle chasing.
四边形 OAPB 是一个风筝形，有两个直角，对称轴是 OP。识别出这个风筝形可以省时间：任何给出同一点引两条切线的题目，都可以用对称性和勾股定理解决，而不必做一长串角度追踪。
A useful numerical pattern: if the radius is 6 cm and OP is 10 cm, then the tangent length is 8 cm, because the right-angled triangle with legs 6 cm and 8 cm has hypotenuse 10 cm.
一个实用的数值规律：若半径为 6 cm，OP 为 10 cm，则切线长为 8 cm，因为直角边为 6 cm 与 8 cm 的直角三角形斜边为 10 cm。
9. Theorem 7: The Alternate Segment Theorem | 定理七：弦切角定理
The angle between a tangent and a chord at the point of contact is equal to the angle in the alternate segment. Write "alternate segment theorem". This is the theorem students most often misapply, so the diagram must be read carefully.
切线与弦在切点处所成的角，等于该弦在另一侧弓形中所对的圆周角。书写理由是 "alternate segment theorem"。这是学生最容易用错的定理，因此必须仔细读图。
In detail: let TA be a tangent at A and AB a chord. The angle TAB (between the tangent and the chord) equals any angle subtended by chord AB at the circumference on the other side of AB. The "alternate segment" is the segment on the opposite side of the chord from the angle you are comparing with.
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