📚 Circle Theorems for IGCSE Maths: The 8 Theorems You Must Master | IGCSE 数学：圆定理八大必考定理与解题技巧
Circle theorems are one of the most predictable topics in IGCSE Mathematics. They appear in both non-calculator and calculator papers, often as multi-step angle reasoning questions. Students who can spot the theorem hidden in a diagram usually gain marks quickly, while those who try to memorise without visual clues often lose easy marks.
圆定理是 IGCSE 数学中出现频率最高的考点之一。无论是非计算器卷还是计算器卷，圆定理常以多步角度推理题的形式出现。能够快速从图形中识别隐藏定理的学生通常能迅速拿分，而只靠死记硬背、不会看图的学生则容易在简单题上失分。
This guide covers the eight circle theorems most commonly examined in IGCSE Mathematics, how to recognise them from diagrams, common mistakes to avoid, and worked exam-style examples.
本指南涵盖 IGCSE 数学中最常考的八大圆定理、如何从图形中识别它们、需要避免的常见错误，以及真题风格的例题精讲。
1. Why Circle Theorems Matter | 为什么圆定理重要
In IGCSE exams, circle theorem questions usually ask you to 'find the value of x' or 'show that angle ABC is 90°'. The marks come from writing clear angle statements and naming the theorem you used. You do not need to prove the theorem from first principles unless the question says so.
在 IGCSE 考试中，圆定理题通常要求你"求 x 的值"或"证明 ∠ABC 是 90°"。得分点在于写出清晰的角度表达式，并注明所使用的定理名称。除非题目明确要求，否则你不需要从基本原理去证明定理本身。
The topic is short, closed and repeatable. Once you can name the eight theorems and recognise their diagram "fingerprints", you can bank marks in almost every past paper. A typical question is worth 3 to 6 marks, and the method marks are awarded for a correct first step even if your final angle is wrong.
这个考点范围封闭、重复率高。一旦你能说全八大定理并识别它们在图中的"指纹"，几乎每份真题都能稳稳拿分。一道典型题值 3 到 6 分，即使最终角度算错，只要第一步推理正确，方法分（M 分）依然可以拿到。
- Exam boards tested: Edexcel IGCSE Mathematics A/B, Cambridge IGCSE 0580 and 0980, Oxford AQA IGCSE, and most national equivalents. ｜ 涉及考局：Edexcel IGCSE 数学 A/B、剑桥 IGCSE 0580 与 0980、牛津 AQA IGCSE，以及多数同类课程。
- Where it appears: non-calculator paper, calculator paper, and geometry sections of the extended tier. ｜ 出现位置：非计算器卷、计算器卷，以及拓展层级的几何板块。
- Typical difficulty: 3-6 marks, multi-step, sometimes combined with trigonometry or Pythagoras. ｜ 常见难度：3-6 分，多步推理，有时与三角比或勾股定理结合。
- Reward per minute: high. It is one of the fastest topics to revise before an exam. ｜ 单位时间回报：很高，是考前复习性价比最高的板块之一。
2. The Language of Angle Reasoning | 角度推理的规范表达
Marks are lost more often through poor notation than through poor geometry. Examiners need to see which angle you mean. Always use three letters with the vertex in the middle: ∠ABC means the angle at B, between BA and BC. Writing "angle B" is risky when more than one angle meets at B.
丢分的原因往往是书写不规范，而不是几何不会。阅卷人必须看清你指的是哪个角。角度一律用三个字母表示，顶点放中间：∠ABC 表示顶点在 B、由 BA 与 BC 夹成的角。当 B 点汇聚多个角时，只写"角 B"非常危险。
Every statement needs a reason in brackets. The reason is a fixed phrase - learn it word for word. "Angles in a triangle sum to 180°" is a reason; "because it's a triangle" is not.
每一个结论后面都要跟括号注明理由，而理由是固定说法，请逐字背诵。"三角形内角和为 180°"是理由，"因为它是三角形"不算理由。
| English reason | 中文理由 | When to use |
| Angle in a semicircle is 90° | 直径所对的圆周角是 90° | Triangle side is a diameter |
| Angles in the same segment are equal | 同弧所对的圆周角相等 | Two angles stand on the same chord |
| Angle at the centre is twice the angle at the circumference | 圆心角是圆周角的两倍 | O is marked, both angles share a chord |
| Opposite angles of a cyclic quadrilateral sum to 180° | 圆内接四边形对角互补 | Four points on the circle form a quadrilateral |
| Alternate segment theorem | 弦切角定理 | A tangent meets a chord at the point of contact |
The safest habit is to write one line per step: angle value, then reason in brackets, then move on. Multi-step questions are marked step by step, so a tidy column of lines earns more than a single crowded sentence.
最稳妥的习惯是"一步一行"：先写角度值，再在括号里写理由，然后继续下一步。多步题是按步骤给分的，整齐的分行作答比把所有内容挤在一句话里更容易得分。
3. Theorem 1: Angle in a Semicircle | 定理一：直径所对圆周角是直角
is any other point on the circumference, then the angle at C is a right angle. In symbols, ∠ACB = 90° whenever AB is a diameter.
若 AB 是圆的直径，C 是圆周上异于 A、B 的任意一点，则 C 处的圆周角是直角。用符号表示：只要 AB 是直径，就有 ∠ACB = 90°。
AB 是直径 ⇒ ∠ACB = 90°
join O to C. OA, OB and OC are all radii, so triangles OAC and OBC are both isosceles. If ∠OAC = α and ∠OBC = β, then ∠OCA = α and ∠OCB = β. The triangle ABC gives 2α + 2β = 180°, so α + β = 90°, which is exactly ∠ACB.
证明思路：连接 OC。OA、OB、OC 都是半径，因此三角形 OAC 与 OBC 都是等腰三角形。设 ∠OAC = α，∠OBC = β，则 ∠OCA = α，∠OCB = β。在三角形 ABC 中有 2α + 2β = 180°，故 α + β = 90°，这正是 ∠ACB 的值。
2α + 2β = 180° ⇒ α + β = 90°
How to recognise it: look for a triangle whose longest side is drawn straight through the centre, or where the side is labelled "diameter". The right-angle mark is usually not printed - that is your job.
如何识别：找那种最长边笔直穿过圆心的三角形，或者题目直接标注该边为"直径"。题目通常不会画出直角符号，这正是你要补上的结论。
- Mistake: assuming the angle is 90° when the side is only a chord. ｜ 常见错误：把弦误当直径，直接判定 90°。
- Mistake: putting the right angle at the wrong vertex. The 90° is always at C, opposite the diameter. ｜ 常见错误：直角放错顶点。90° 一定在 C 处，也就是直径的对角。
- Exam tip: this theorem often appears as the final step, after you have found the other two angles of the triangle. ｜ 应试技巧：该定理常作为最后一步出现，等你先求出三角形的另外两角之后再用。
4. Theorem 2: Angles in the Same Segment | 定理二：同弧所对圆周角相等
ord, are equal. If A, B, C and D lie on a circle and C, D are on the same side of AB, then ∠ACB = ∠ADB.
同一条弦在同一侧所对的圆周角相等。若 A、B、C、D 四点在圆上，且 C、D 位于 AB 同侧，则 ∠ACB = ∠ADB。
∠ACB = ∠ADB（同弧 AB，同侧）
This is the "bow-tie" or "butterfly" shape: two triangles sharing a common base, with their apexes on the circle. In a diagram it looks like a chord with two arms reaching up to two different points on the arc.
这就是经典的"领结形"或"蝴蝶形"：两个三角形共用一条底边，顶点都在圆上。图中表现为一条弦向上伸出两组"手臂"，分别连到弧上的两个不同点。
The converse is also examinable: if two angles standing on the same side of a line are equal, then the four points are concyclic. This is a neat way to prove that four points lie on one circle.
它的逆定理同样可能考到：若一条线段同侧的两个角相等，则这四个点共圆。这是证明四点共圆的漂亮方法。
- Recognise it: a common chord, two apex points, equal marks or the word "same segment". ｜ 识别特征：共用一条弦、两个顶点、图上标记相等角或出现"same segment"字样。
- Mistake: using the theorem when the two angles are on opposite sides of the chord - there it becomes a cyclic quadrilateral instead. ｜ 常见错误：两角位于弦的两侧时误用此定理，此时应改用圆内接四边形。
- Mistake: assuming equality because the shape looks symmetrical. ｜ 常见错误：仅凭图形看起来对称就断定相等。
5. Theorem 3: Angle at the Centre | 定理三：圆心角等于圆周角的两倍
The angle subtended by an arc at the centre of a circle is twice the angle subtended by the same arc at the circumference, provided both stand on the same arc.
同一条弧所对的圆心角，等于它所对圆周角的两倍（前提是两者所对的弧相同）。
∠AOB = 2 × ∠ACB
e. If the centre angle shown is reflex, it is still double the circumference angle: reflex ∠AOB = 2 × ∠ACB. Students often find 2x = 360° − reflex and get confused.
遇到优角（大于 180° 的角）时要特别小心。若图中的圆心角是优角，它依然是圆周角的两倍：优角 ∠AOB = 2 × ∠ACB。很多学生在用 360° 减去优角时把自己绕晕。
Worked micro-example: O is the centre, ∠AOB = 130°, C lies on the major arc. Then ∠ACB = 130 ÷ 2 = 65°. If instead ∠ on-reflex angle at O is 96° only if 96° is the angle on the same side. Always sketch and mark the arc you are using.
微型例题：O 为圆心，∠AOB = 130°，C 在优弧上，则 ∠ACB = 130 ÷ 2 = 65°。反过来，若 ∠ACB = 48°，则 O 处对应的优角为 96°。务必画图标出你所使用的那条弧，避免张冠李戴。
- Recognise it: the letter O is marked, and a radius or two is drawn to the vertices. ｜ 识别特征：图中标注了圆心 O，并画出连接顶点的一两条半径。
- Mistake: halving the wrong angle - check whether the given angle is at the centre or the circumference. ｜ 常见错误：除错了对象，先确认已知角是圆心角还是圆周角。
- Exam tip: this theorem is the backbone of cyclic quadrilateral proofs. ｜ 应试技巧：这个定理是圆内接四边形证明的核心工具。
6. Theorem 4: Cyclic Quadrilateral | 定理四：圆内接四边形对角互补
vertices all lie on a circle. Its opposite angles add up to 180°. If ABCD is cyclic, then ∠ABC + ∠ADC = 180° and ∠BAD + ∠BCD = 180°.
圆内接四边形是指四个顶点都在同一个圆上的四边形。它的对角互补，即相加等于 180°。若 ABCD 内接于圆，则 ∠ABC + ∠ADC = 180°，∠BAD + ∠BCD = 180°。
∠A + ∠C = 180°， ∠B + ∠D = 180°
Proof sketch: let the centre be O. The angle at O subtended by chord AB is twice ∠ADB, and the angle at O subtended by the same chord on the other side is twice ∠ACB. These two centre angles add to 360°, so 2∠ADB + 2∠ACB = 360°, giving ∠ADB + ∠ACB = 180°.
证明思路：设圆心为 O。弦 AB 在 O 处所对的角等于 2∠ADB；同一弦在另一侧所对的圆心角等于 2∠ACB。这两个圆心角加起来是 360°，故 2∠ADB + 2∠ACB = 360°，两边除以 2 得 ∠ADB + ∠ACB = 180°。
There is a very useful companion result: the exterior angle of a cyclic quadrilateral equals the interior opposite angle. If you extend side BC beyond C, the exterior angle at C equals ∠BAD.
还有一个极具实用价值的推论：圆内接四边形的外角等于内对角。若把边 BC 延长到 C 之外，则 C 处的外角等于 ∠BAD。
- Recognise it: four labelled points on a circle joined in order; often one angle is given and you must find its opposite. ｜ 识别特征：圆上四个点按顺序连成四边形，常给出一个角要求其对角。
- Mistake: adding adjacent angles instead of opposite ones. ｜ 常见错误：把相邻角当成对角相加。
- Mistake: forgetting that the theorem only works when all four vertices are on the circle. ｜ 常见错误：忘记前提是四个顶点都在圆上。
7. Theorem 5: Alternate Segment Theorem | 定理五：弦切角定理
The angle between a tangent and a chord drawn from the point of contact is equal to the angle in the alternate segment. If TA is a tangent at A and AB is a chord, then ∠TAB equals the angle that any point C on the far arc makes with AB, i.e. ∠ACB.
切线与从切点引出的弦之间的夹角，等于该弦所对的另一个弓形中的圆周角。若 TA 在 A 处与圆相切，AB 是一条弦，则 ∠TAB 等于远弧上任意一点 C 所张的角 ∠ACB。
∠TAB = ∠ACB（C 在另一弓形中）
The word "alternate" is the key. The angle you are comparing with sits in the segment on the opposite side of the chord from the tangent-chord angle. If you pick the angle on the same side, you get a supplementary relationship instead.
"Alternate（交错）"是关键。与你作比较的那个圆周角，必须位于弦的另一侧弓形中，与弦切角分居弦的两边。若选到同侧的角，得到的就是互补关系而不是相等关系。
This is usually the theorem students find hardest to see, so practise spotting the three required elements: a tangent line, a chord from the contact point, and a third point on the circumference.
这是学生最难看出来的一个定理，因此请刻意练习识别三个必备元素：一条切线、一条从切点引出的弦、圆周上的第三个点。
- Recognise it: a straight line touching the circle at a labelled point, with a chord and a triangle inside. ｜ 识别特征：一条直线在标出的点处与圆相切，圆内有弦和三角形。
- Mistake: choosing the angle in the same segment rather than the alternate one. ｜ 常见错误：选错弓形，取成了同侧的角。
- Exam tip: this theorem frequently combines with the tangent-radius rule (Theorem 6) in the same question. ｜ 应试技巧：它与定理六（切线垂直于半径）常在同一题中配合使用。
8. Theorem 6: Tangent Perpendicular to Radius | 定理六：切线垂直于过切点的半径
A tangent touches a circle at exactly one point, and the radius drawn to that point of contact is perpendicular to the tangent. In symbols, if T is the point of contact and O is the centre, then OT ⟂ tangent at T, so the angle is exactly 90°.
切线与圆只有一个公共点；连接圆心与该切点的半径垂直于切线。用符号表示：若 T 为切点、O 为圆心，则 OT ⟂ 过 T 的切线，即夹角恰为 90°。
OT ⟂ tangent at T ⇒ ∠OTP = 90°
This single fact turns many questions into right-angled triangle problems. Once you can mark a 90° angle, you can use Pythagoras or the sine, cosine and tangent ratios. For an external point P with tangent PT and radius r, the right-angled triangle OTP gives:
这一条事实能把大量题目转化为直角三角形问题。一旦标出 90°，你就可以使用勾股定理或正弦、余弦、正切。对于圆外一点 P、切线 PT 与半径 r，直角三角形 OTP 满足：
OP² = OT² + TP² = r² + TP²
A tangent also implies equal angles with any chord only through Theorem 5, but the 90° with the radius is unconditional and always available.
切线只有通过定理五才与弦产生等角关系，但与半径垂直成 90° 这一条是无条件的，任何时候都能使用。
- Recognise it: the phrase "PT is a tangent", or a line that clearly just grazes the circle. ｜ 识别特征：题干出现"PT is a tangent"，或一条明显只与圆"擦边"的直线。
- Mistake: drawing the right angle at the wrong end, i.e. at O instead of at the point of contact. ｜ 常见错误：直角标错位置，标在了 O 处而不是切点处。
- Exam tip: pair this with Pythagoras to find lengths; the answer often needs a square root left as a simplified surd. ｜ 应试技巧：与勾股定理配合求长度，答案常需保留为最简根式。
9. Theorem 7: Two Tangents from an External Point | 定理七：圆外一点引两条切线长度相等
are the points of contact. Moreover, the line OP bisects the angle between the tangents (∠APB) and also bisects the angle at the centre (∠AOB).
从圆外一点 P 引圆的两条切线，两条切线长相等：PA = PB（A、B 为切点）。此外，直线 OP 平分两条切线之间的夹角 ∠APB，同时也平分圆心处的角 ∠AOB。
PA = PB， ∠APO = ∠BPO， ∠AOP = ∠BOP
Why? both share the hypotenuse OP, and OA = OB = r. Congruence gives all three results at once.
原因在于：三角形 OAP 与 OBP 满足 RHS 全等条件——两处切点角都是直角，共用斜边 OP，且 OA = OB = r。全等一次性给出上述三个结论。
The shape OAPB is a kite that is also cyclic, since ∠OAP = ∠OBP = 90° means the opposite angles sum to 180°. This is a neat link back to Theorem 4.
四边形 OAPB 是一个"风筝形"，同时它也是圆内接
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