📚 Circle Theorems for IGCSE Mathematics | IGCSE 数学圆定理核心总结
Circle theorems are a key part of the IGCSE Mathematics syllabus. They describe the relationships between angles, lines and lengths inside a circle. Mastering these theorems will help you solve geometry problems quickly and confidently, especially in the non-calculator paper. This guide covers every circle theorem you need, shows the exact reason statements examiners expect, and flags the mistakes that cost the most marks.
圆定理是 IGCSE 数学大纲中的重点内容。它们描述了圆中角、线与长度之间的关系。掌握这些定理能帮助你快速且自信地解决几何题，尤其是在非计算器卷中。本文覆盖你需要的全部圆定理，给出考官认可的标准理由写法，并指出最容易失分的陷阱。
1. Basic Circle Vocabulary and Notation | 圆的基本词汇与记号
Before you can quote a theorem, you must be able to name the parts of a circle correctly. The centre O is the point equidistant from every point on the circle, and that fixed distance is the radius r. The circumference is the boundary itself. A chord is a straight line segment joining two points on the circumference, and a diameter is simply the longest chord, one that passes through the centre. An arc is a part of the circumference; the minor arc is the shorter one and the major arc is the longer one.
在引用定理之前，你必须能正确说出圆的各部分名称。圆心 O 是与圆上每一点距离都相等的点，这个固定距离就是半径 r。圆周是圆本身的边界。弦是连接圆周上两点的直线段，而直径就是最长的弦，即经过圆心的弦。弧是圆周的一部分；劣弧是较短的那段，优弧是较长的那段。
The verb subtend is used constantly in mark schemes. A chord AB subtends an angle at a point P if P is the vertex of the angle and the two rays of that angle pass through A and B. Similarly, an arc or a chord can subtend an angle at the centre O. Getting this language right is half the battle in a proof question.
subtend（对……张角）这个词在评分标准中反复出现。如果点 P 是角的顶点，且角的两条射线经过 A 和 B，就说弦 AB 在点 P 处张成一个角。同样，一段弧或一条弦也可以在圆心 O 处张成一个角。把这套语言说准确，证明题就成功了一半。
- Centre 圆心 / Radius 半径 / Diameter 直径
- Circumference 圆周 / Arc 弧 / Chord 弦
- Minor segment 劣弓形 / Major segment 优弓形
- Tangent 切线（与圆只有一个公共点）/ Secant 割线（与圆有两个公共点）
- Cyclic quadrilateral 圆内接四边形（四个顶点都在圆上）
2. Angle at the Centre Theorem | 圆心角定理
he circle, O is the centre and C is another point on the major arc, then the central angle ∠AOB equals 2 × ∠ACB.
同一条弧在圆心处所对的角，等于它在圆周上所对的角的两倍。如果 A 和 B 是圆上两点，O 是圆心，C 是优弧上的另一点，那么圆心角 ∠AOB 等于 2 × ∠ACB。
∠AOB = 2 × ∠ACB
e the same chord AB. Identify the chord first, then apply the ratio 2 : 1. For example, if ∠ACB = 34°, then ∠AOB = 68°.
这个定理经常在图中考查，其中圆心角和圆周角共用同一条弦 AB。先找到公共弦，再应用 2 : 1 的比例关系。例如，若 ∠ACB = 34°，则 ∠AOB = 68°。
lies on the minor arc AB, the angle you need at the centre is the reflex angle, so 360° − ∠AOB = 2 × ∠ACB. Always check where the vertex C sits before you double or halve anything.
有一个微妙之处在于该用哪个圆心角。当 C 位于劣弧 AB 上时，你在圆心需要的角是优角，即 360° − ∠AOB = 2 × ∠ACB。所以在你做任何"加倍"或"减半"之前，先确认顶点 C 的位置。
A useful consequence: if ∠AOB = 180°, then ∠ACB = 90°, which gives the next theorem directly.
一个有用的推论：若 ∠AOB = 180°，则 ∠ACB = 90°，这直接得到下一个定理。
3. Angle in a Semicircle | 半圆上的圆周角
is a right angle. If AB is a diameter of the circle and P lies on the circumference, then ∠APB = 90°.
直径所对的圆周角是直角。如果 AB 是圆的直径，点 P 在圆周上，那么 ∠APB = 90°。
AB is a diameter ⇒ ∠APB = 90°
ful and is often the point of a question: if a triangle inscribed in a circle has a right angle at P, then the side opposite P, namely AB, must be a diameter, so the midpoint of AB is the centre of the circle.
它其实就是圆心角定理的特例：直径在圆心 O 处张成 180°，而 180° ÷ 2 = 90°。它的逆定理同样有力，而且常常正是题目的考点：如果圆内接三角形在 P 处为直角，那么 P 所对的边 AB 必是直径，因此 AB 的中点就是圆心。
Typical phrasing in an exam: "A, B and C lie on a circle with centre O. Angle ABC = 90°. Explain why AC is a diameter." hrough three points by locating the midpoint of the hypotenuse of a right-angled triangle.
考试中的典型问法："A、B、C 三点在圆心为 O 的圆上，∠ABC = 90°，请说明 AC 是直径。"所需理由就是半圆圆周角定理的逆定理。在坐标几何中，这一定理还能让你通过定位直角三角形斜边的中点，来求出过三点的圆。
4. Angles in the Same Segment | 同弧上的圆周角
If A, B, C and D are points on a circle and C and D lie on the same side of chord AB, then ∠ACB = ∠ADB.
同一弓形内的圆周角相等。如果 A、B、C、D 四点共圆，且 C 与 D 位于弦 AB 的同一侧，那么 ∠ACB = ∠ADB。
∠ACB = ∠ADB (same chord AB, same side)
Both angles subtend the same arc AB, so both equal half of the central angle ∠AOB. This is the theorem that turns an awkward angle into a friendly one: you look for a "bow-tie" or "butterfly" shape formed by two triangles sharing the same chord.
这两个角都对着同一段弧 AB，因此都等于圆心角 ∠AOB 的一半。这个定理能把一个不好求的角转换成一个好求的角：你要在图中寻找两个三角形共用同一条弦所形成的"蝴蝶形"结构。
Watch the side condition carefully. If C and D are on opposite sides of AB, the two angles are not equal; instead they add up to 180°, because ABCD becomes a cyclic quadrilateral. This is exactly why the next theorem exists.
要特别注意"同侧"这个条件。如果 C 与 D 位于 AB 的两侧，两个角并不相等，而是互补为 180°，因为此时 ABCD 构成圆内接四边形。这正是下一个定理存在的原因。
5. Cyclic Quadrilaterals | 圆内接四边形
A cyclic quadrilateral is a four-sided shape whose four vertices all lie on a circle. Its opposite angles are supplementary, that is, they add up to 180°.
圆内接四边形是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°。
∠A + ∠C = 180° and ∠B + ∠D = 180°
ral equals the interior opposite angle. If side AB is extended beyond B to a point E, then ∠CBE = ∠ADC. This follows immediately from the supplementary angles rule, since ∠ABC + ∠CBE = 180° on a straight line.
与之相关且极常考的一条性质是外角定理：圆内接四边形的外角等于它的内对角。如果把边 AB 延长到 B 之外的 E 点，则 ∠CBE = ∠ADC。这可由对角互补直接推出，因为 ∠ABC + ∠CBE = 180°（同一直线上的邻补角）。
The converse also matters: if a pair of opposite angles of a quadrilateral add up to 180°, then the four points are concyclic, meaning they lie on a single circle. This is how you prove four points are concyclic without drawing anything extra.
逆定理同样重要：如果四边形的一组对角之和为 180°，那么这四个点共圆。这就是在不额外作图的情况下证明四点共圆的方法。
6. Tangent and Radius: The 90° Rule | 切线与半径的垂直关系
to the point of contact. If the line PT touches the circle at T and O is the centre, then ∠OTP = 90°.
圆的切线垂直于过切点的半径。若直线 PT 在 T 点与圆相切，O 为圆心，则 ∠OTP = 90°。
OT ⊥ PT ⇒ ∠OTP = 90°
The converse is a standard method for proving a line is a tangent: if a line is perpendicular to a radius at a point on the circumference, then it is a tangent at that point. Another equivalent form states that the perpendicular distance from the centre to a tangent equals the radius.
逆定理是证明某直线为切线的标准方法：若一条直线在圆周上某点处垂直于半径，则它在该点与圆相切。另一个等价说法是：圆心到切线的垂直距离等于半径。
In problem solving this rule almost always creates a right-angled triangle, so Pythagoras' ic trigonometry come straight into play. Given OT = 5 cm and PT = 12 cm, the distance OP = √5² + 12² = √169 = 13 cm.
解题时这条定理几乎总会构造出一个直角三角形，因此勾股定理与基本三角比立刻就能用上。已知 OT = 5 cm、PT = 12 cm，则 OP = √5² + 12² = √169 = 13 cm。
7. Two Tangents from an External Point | 圆外一点引出的两条切线
th centre O, then PA = PB. The two tangent lengths from the same external point are equal.
从圆外一点 P 向圆心为 O 的圆作两条切线 PA 和 PB，则 PA = PB。即从同一外点引出的两条切线长度相等。
PA = PB, ∠APO = ∠BPO, OP bisects ∠APB
e proof uses congruent triangles. Triangles OAP and OBP are congruent by RHS: OA = OB = r, ∠OAP = ∠OBP = 90° because both are tangent radii, and OP is common. It follows that PA = PB, ∠APO = ∠BPO, and ∠AOP = ∠BOP.
证明用到全等三角形。△OAP 与 △OBP 由 RHS 判定全等：OA = OB = r，∠OAP = ∠OBP = 90°（均为切线垂直于半径），OP 为公共边。由此可得 PA = PB、∠APO = ∠BPO、∠AOP = ∠BOP。
A further consequence is that OP is the perpendicular bisector of the chord AB. Since △OAB is isosceles with OA = OB, and OP bisects the apex angle ∠ se AB at right angles. This single configuration unlocks a large family of exam questions.
进一步的推论是：OP 是弦 AB 的垂直平分线。因为 △OAB 中 OA = OB 是等腰三角形，而 OP 平分顶角 ∠AOB，所以它必然垂直平分底边 AB。这一个图形结构能解锁一大类考题。
8. Alternate Segment Theorem | 切线-弦夹角定理（交替弓形定理）
The angle between a tangent and a chord at the point of contact equals the angle in the alternate segment. If the tangent at A touches the circle and AB is a chord, then the angle between the tangent and AB equals the angle subtended by AB at any point C on the circumference on the other side of AB.
切线与弦在切点处的夹角，等于该弦在另一侧弓形中所对的圆周角。若圆在 A 处的切线与弦 AB 构成夹角，则该夹角等于 AB 在弦另一侧圆周上任一点 C 处所张的角。
angle between tangent and AB = ∠ACB (alternate segment)
Students often find this the trickiest theorem, so here is a reliable procedure. First mark the point of contact and identify the chord. Then find the triangle that sits on the other side of that chord. The angle at the third vertex of that triangle is the one that equals your tangent-chord angle.
学生通常觉得这条定理最难，这里给出一个可靠的操作步骤。先在切点处做标记，确认是哪条弦。然后找出位于这条弦另一侧的那个三角形。该三角形第三个顶点处的角，就等于你要求的切线与弦的夹角。
Remember that there are two tangent-chord angles at the point of contact, one on each side of the chord, and each matches the angle in its own alternate segment. A quick check: the two angles in the two segments are supplementary, just as the two tangent-chord angles are.
注意在切点处有两个切线与弦的夹角，分别位于弦的两侧，各自对应自己那一侧的弓形角。一个快速验证：两侧弓形中的圆周角互补，两个切线与弦的夹角同样互补。
9. Chord Properties: Perpendicular from Centre and Intersecting Chords | 弦的性质：垂径定理与相交弦定理
The perpendicular from the centre of a circle to a chord bisects the chord. If OM ⊥ o bisects the arc AB. Conversely, the perpendicular bisector of a chord passes through the centre, which is a practical way to find the centre of a circle given three points on it.
从圆心向弦作垂线，垂足平分这条弦。若 OM ⊥ AB，O 为圆心，M 在弦 AB 上，则 AM = MB。同一条直线还平分弧 AB。反过来，弦的垂直平分线经过圆心，这正是只给定圆上三点时寻找圆心的实用方法。
OM ⊥ AB ⇒ AM = MB
f a circle of radius r, then a² + d² = r². For example, a chord of length 16 cm in a circle of radius 10 cm is 6 cm from the centre, since 8² + 6² = 10².
结合勾股定理，可得到标准的半径-弦关系式。若长为 2a 的弦到半径为 r 的圆心的垂直距离为 d，则 a² + d² = r²。例如，半径 10 cm 的圆中一条长 16 cm 的弦，距圆心 6 cm，因为 8² + 6² = 10²。
is also the tangent-secant version: PA² = PB × PC. Check your specification before relying on these, as they appear more often in Additional or Further Mathematics.
作为拓展内容，相交弦定理指出：当两条弦 AB 与 CD 在圆内一点 P 相交时，AP × PB = CP × PD。还有切割线定理的版本：PA² = PB × PC。使用前请先核对你的考纲，因为它们更多出现在 Additional 或 Further Mathematics 中。
10. How to Write a Proof: Reason Statements | 证明题的书写与理由规范
IGCSE geometry questions award marks for reasons, not just for the correct angle. Every step of your working needs a named theorem. In general, write one line of equation followed by the reason in brackets, and never skip the reason hoping the examiner will infer it.
IGCSE 几何题是按"理由"给分的，而不仅仅是看答案是否正确。你的每一步推理都要写出所用定理的名称。一般写法是：先写一行算式，后面用括号写出理由，绝不要省掉理由指望考官替你推断。
- Angle at the centre is twice the angle at the circumference 圆心角是圆周角的两倍
- Angle in a semicircle is 90° 半圆所对的圆周角是 90°
- Angles in the same segment are equal 同弓形内的圆周角相等
- Opposite angles of a cyclic quadrilateral sum to 180° 圆内接四边形对角互补
- Exterior angle of a cyclic quadrilateral equals the interior opposite angle 圆内接四边形外角等于内对角
- Tangent is perpendicular to the radius at the point of contact 切线垂直于过切点的半径
- Tangents from an external point are equal in length 圆外一点引出的两条切线长度相等
- Alternate segment theorem 交替弓形定理
- Perpendicular from the centre to a chord bisects the chord 垂径定理
- Base angles of an isosceles triangle are equal 等腰三角形两底角相等
- Angles in a triangle sum to 180° 三角形内角和为 180°
Notice the last two entries. Many circle theorem questions finish with an ordinary triangle or isosceles triangle argument, so you should be ready to switch between circle facts and basic angle facts in the same proof.
注意最后两条。许多圆定理题的最后一步要用到普通三角形或等腰三角形的结论，所以你要能在同一道证明中自如地在圆的性质与基本角的性质之间切换。
11. Common Mistakes and a Self-Check Routine | 常见易错点与自查清单
The most frequent error is confusing the centre theorem with the same-segment theorem, which produces answers that are out by a factor of two. Another common slip is assuming that angles on opposite sides of the same chord are equal, when in fact they are supplementary.
最常见的错误是把圆心角定理与同弓形定理搞混，导致答案差了两倍。另一个常见失误是以为同一条弦两侧的圆周角相等，而实际上它们互补。
- Halving when you should double, or doubling when you should halve 该减半时却加倍，或该加倍时却减半
- Using the wrong central angle when the point lies on the minor arc 点在劣弧上时用错了圆心角
- Assuming a line is a tangent just because it looks like one 仅凭"看起来像"就认定某直线是切线
- Assuming an angle is 90° from the diagram rather than from a diameter 从图上看像直角就当作 90°，而非由直径推出
- Quoting a theorem without checking the configuration is valid 引用定理时未检查图形是否满足条件
- Leaving answers as unsimplified fractions or forgetting the degree symbol 答案未化简或漏写度数符号
Build a habit of a ten-second check after each answer. Ask yourself: which chord or arc does this angle subtend, is the vertex on the circumference or at the centre, and which theorem did I actually use? Then verify that your angles in a triangle still sum to 180°.
养成每得出一个答案后花十秒自查的习惯。问自己：这个角对着的是哪条弦或哪段弧？顶点在圆周上还是在圆心？我实际用到了哪条定理？最后再验证三角形内角和是否仍为 180°。
12. Worked Example and One-Page Summary | 真题例题与一页速查表
ABC = 32°. The tangent at B meets the line AC produced at T. Find angle BAC and angle ABT.
来看一道标准题。A、B、C 三点在圆心为 O 的圆上，AB 是直径，∠ABC = 32°。过 B 的切线与 AC 的延长线交于 T。求 ∠BAC 与 ∠ABT。
Step one: since AB is a diameter, ∠ACB = 90° by the angle in a semicircle theorem. Step two: in triangle ABC, ∠ a triangle sum to 180°. Step three: the tangent at B is perpendicular to the radius OB, so ∠ ternate segment theorem directly: the angle between the tangent BT and the chord AB equals ∠ACB, and since ∠ACB = 90°... in that case use the angle between tangent BT and chord BC, which equals ∠BAC = 58°.
第一步：因为 AB 是直径，由半圆圆周角定理得 ∠ACB = 90°。第二步：在三角形 ABC 中，∠BAC = 180° − 90° − 32° = 58°，依据是三角形内角和为 180°。第三步：过 B 的切线垂直于半径 OB，所以 ∠ABT = 90° − 32° = 58°。也可以直接引用交替弓形定理：切线与弦所成的角等于另一侧弓形角，从而同样得到 58°。
| Theorem 定理 | Key result 核心结论 | Trigger in the diagram 图中触发条件 |
| Angle at the centre 圆心角定理 | ∠AOB = 2 × ∠ACB | style="border:1px solid #ccc;padding
广告 / Sponsored 【现货】剑桥国际高中教材 Cambridge IGCSE & O Level Additional Mathemati… 数学 备考教辅 · 正版现货 · 与本文内容相关 查看详情 / View on Taobao参考价 ¥198.0更多咨询请联系16621398022（同微信） 备考资讯每日更新，关注不错过 考纲变化 · 科目分析 · 学习方法  在微信里长按二维码识别关注 或截图后用「扫一扫 → 相册」选择本图 公众号：tutorhao |