📚 IGCSE Circle Theorems: Complete Topic T-4-1028 | IGCSE 圆定理：完整专题 T-4-1028
Circle theorems are a central part of IGCSE geometry. They describe angle and length relationships involving circles, chords, tangents, and cyclic quadrilaterals. Mastering these rules allows you to find unknown angles quickly and to justify each step in a proof-style question. This topic page, T-4-1028, collects every circle theorem you need for Cambridge IGCSE Mathematics and Edexcel IGCSE, together with the exact wording examiners expect in your reasons.
圆定理是 IGCSE 几何的核心内容。它们描述圆、弦、切线和圆内接四边形之间的角度与长度关系。掌握这些规则可以帮助你快速求出未知角，并在几何证明题中清楚地写出每一步依据。本篇专题 T-4-1028 汇总了剑桥 IGCSE 数学与 Edexcel IGCSE 所需的全部圆定理，并附上考官期待看到的规范理由表述。
1. What Circle Theorems Are and How They Are Examined | 圆定理是什么以及考试如何考查
There are seven standard circle theorems in the IGCSE specification. Four of them are angle theorems (centre, same segment, semicircle, cyclic quadrilateral), two concern tangents (tangent-radius and alternate segment), and one concerns chords (perpendicular from the centre). Questions normally appear on the non-calculator or calculator geometry paper, and they are worth between three and six marks.
IGCSE 考纲中共有七条标准圆定理。其中四条是角度定理（圆心角、同弓形、半圆、圆内接四边形），两条与切线有关（切线垂直于半径、弦切角），一条与弦有关（垂径定理）。这类题通常出现在几何部分，分值在 3 到 6 分之间。
Marks are not only for the final angle. Each line of working must carry a reason, written in words, and this reasoning mark is awarded separately. A correct answer with no reasons can lose half the marks, so learn the phrases as carefully as you learn the numbers.
分数并不只看最后的答案。每一步推导都必须写出文字理由，而理由分是单独给分的。答案正确但完全不写理由，可能会丢掉一半的分数，所以背理由表述要和算数一样认真。
- Chord - a straight line joining two points on the circumference. | 弦：连结圆上两点的线段。
- Tangent - a line that touches the circle at exactly one point. | 切线：与圆恰好只有一个公共点的直线。
- Arc - a part of the circumference. | 弧：圆周的一部分。
- Segment - the region between a chord and an arc. | 弓形：弦与弧之间的区域。
- Subtend - an angle stands on, or is supported by, an arc. | 对所：角所站立的那条弧。
- Cyclic quadrilateral - a four-sided shape whose vertices all lie on one circle. | 圆内接四边形：四个顶点都在同一个圆上的四边形。
- Concyclic - points that all lie on the same circle. | 共圆：所有点位于同一个圆上。
Always mark the centre O, the radius, and any tangents on the diagram before you start. A diagram that is clearly labelled turns a six-mark question into a two-minute question.
动笔之前，先在图上标出圆心 O、半径以及所有切线。标清图形能让 6 分题变成两分钟就能做完的题。
2. Theorem 1: Angle at the Centre | 定理一：圆心角定理
The first core rule states that the angle subtended by an arc at the centre of a circle is exactly twice the angle subtended by the same arc at the circumference. This works for any arc, whether minor or major, as long as the two angles stand on the same arc.
第一条核心规则是：同一条弧所对的圆心角等于该弧所对圆周角的两倍。无论这条弧是优弧还是劣弧，只要两个角站在同一条弧上，这个关系都成立。
∠AOC = 2 × ∠ABC
In a diagram, if O is the centre and A, B, C are points on the circle, angle AOC at the centre is twice angle ABC at the circumference. You can use this to find either angle when the other is known, but be careful to identify the same arc.
在图中，如果 O 是圆心，A、B、C 是圆上的点，那么圆心角 ∠AOC 是圆周角 ∠ABC 的两倍。你可以利用这个关系由其中一个角求另一个角，但必须注意确认两个角所对的是同一条弧。
flex angle, measured on the far side of O, and the two angles add to 360°. For example, if ∠ABC = 47°, then the minor angle ∠AOC = 94° and the reflex angle ∠AOC = 2 × (180° − 47°) = 266°, and 94° + 266° = 360°.
常见的陷阱是优角。如果圆周角是钝角，那么同一条弧所对的圆心角其实是优角，也就是从 O 的另一侧量得的那个角，两个角之和为 360°。例如 ∠ABC = 47° 时，劣角 ∠AOC = 94°，优角 ∠AOC = 2 × (180° − 47°) = 266°，而 94° + 266° = 360°。
Standard reason: "the angle at the centre is twice the angle at the circumference" or "angle at centre = 2 × angle at circumference". Write one of these, not just "because of the theorem".
规范理由写法："the angle at the centre is twice the angle at the circumference"（圆心角等于圆周角的两倍）。请写完整句子，不要只写"根据定理"。
3. Theorem 2: Angles in the Same Segment | 定理二：同弓形上的圆周角相等
All angles in the same segment, standing on the same chord, are equal. This means that if several angles are drawn from the same chord to different points on the same arc, their sizes are identical.
位于同一弓形内、由同一条弦所对的所有圆周角都相等。也就是说，如果一条弦对应的几个圆周角指向同一条弧上的不同点，它们的大小完全相同。
∠APB = ∠AQB
This theorem is very useful in cyclic diagrams where a chord appears to support two or more angles at the circumference. Look for the same base chord and the same side of the chord before stating equality.
这个定理在圆内接图形中非常有用，当一条弦支撑两个或更多圆周角时就会出现。使用前要先找到同一条底弦，并确认这些角位于弦的同侧。
The theorem follows directly from Theorem 1. Both ∠APB and ∠AQB are subtended by the arc AB, so both are equal to half of the same angle at the centre. Seeing the link between the two theorems makes it much harder to forget either of them.
这条定理其实是定理一的直接推论。∠APB 和 ∠AQB 都由弧 AB 所对，所以它们都等于同一个圆心角的一半。理解了这两条定理之间的联系，就很难再忘记其中任何一条。
The converse is also examinable: if two points P and Q lie on the same side of AB and ∠APB = ∠ then A, B, P and Q are concyclic. This is how you prove that four points lie on a circle.
它的逆定理同样可能考到：若 P、Q 在 AB 同侧且 ∠APB = ∠AQB，则 A、B、P、Q 四点共圆。这就是证明四点共圆的标准方法。
Standard reason: "angles in the same segment are equal". Never write "same arc" when you mean "same segment" - the wording matters to the examiner.
规范理由写法："angles in the same segment are equal"（同弓形上的圆周角相等）。不要把 segment 写成 arc，用词在评分中很重要。
4. Theorem 3: Angle in a Semicircle | 定理三：半圆上的圆周角是直角
An angle inscribed in a semicircle, that is an angle subtended by a diameter, is always a right angle of 90°. Because the diameter subtends 180° at the centre, the angle at the circumference must be half of that, namely 90°.
半圆内接的角，也就是直径所对的圆周角，永远是 90° 直角。因为直径在圆心处所对的角是 180°，圆周角必然是它的一半，即 90°。
If AB is a diameter, then ∠APB = 90°
This is the theorem behind many right-angle proofs. If a question asks you to show that a triangle is right-angled, look for a diameter, or prove that one side must be a diameter by showing the angle opposite it is 90°.
这条定理是许多直角证明的基础。题目要求证明三角形是直角三角形时，先找直径；反过来，若某个角是 90°，则它对面的那条边就是直径。
The converse is used just as often: if the angle subtended by a chord at the circumference is 90°, then that chord is a diameter, so its midpoint is the centre of the circle. This is the standard way to locate the centre of a circle from a right-angled construction.
逆定理同样常用：如果一条弦所对的圆周角是 90°，那么这条弦就是直径，其中点就是圆心。这是通过直角构造确定圆心的标准方法。
Do not confuse this with "the angle between a tangent and a chord". The 90° in this theorem sits on the circumference; the 90° in the tangent theorem sits between a tangent and a radius. Read the diagram carefully before quoting a reason.
不要把它和"切线与弦的夹角"混淆。本条定理中的 90° 在圆周上；切线定理中的 90° 位于切线与半径之间。引用理由前请仔细看图。
Standard reason: "the angle in a semicircle is 90°" or "angle subtended by a diameter is 90°".
规范理由写法："the angle in a semicircle is 90°" 或 "angle subtended by a diameter is 90°"。
5. Theorem 4: Cyclic Quadrilaterals | 定理四：圆内接四边形
s adds up to 180°. If the vertices A, B, C and D all lie on one circle in that order, then ∠A + ∠C = 180° and ∠B + ∠D = 180°.
圆内接四边形中，每组对角之和都等于 180°。如果顶点 A、B、C、D 按顺序都在同一个圆上，则 ∠A + ∠C = 180°，∠B + ∠D = 180°。
∠A + ∠C = 180° and ∠B + ∠D = 180°
There is a second, equally important form of this theorem: an exterior angle of a cyclic quadrilateral equals the interior opposite angle. Both forms come from the same result, and either can be quoted depending on which angle the question gives you.
这条定理还有另一个同样重要的形式：圆内接四边形的一个外角等于它的内对角。两种形式来自同一个结论，具体引用哪一种取决于题目给出的是哪个角。
h: join B and D to the centre O. The reflex angle at the centre standing on arc BCD is 2 × ∠A, and the angle at the centre on arc BAD is 2 × ∠C. These two centre angles sum to 360°, so ∠A + ∠C = 180°. Knowing this proof earns full marks if the question says "prove that".
证明思路：连结 B、D 与圆心 O。弧 BCD 所对的优角圆心角是 2 × ∠A，弧 BAD 所对的圆心角是 2 × ∠C。这两个圆心角之和为 360°，所以 ∠A + ∠C = 180°。若题目要求"证明"，写出这个证明过程可以拿满分。
The converse matters too: if opposite angles of a quadrilateral sum to 180°, the quadrilateral is cyclic. Many exam questions first build a quadrilateral from two triangles and then ask you to prove the four points are concyclic.
逆定理也很重要：若一个四边形的对角之和为 180°，则它内接于圆。许多考题先由两个三角形拼出一个四边形，再要求你证明四点共圆。
Standard reason: "opposite angles of a cyclic quadrilateral sum to 180°" or "exterior angle of a cyclic quadrilateral equals the interior opposite angle".
规范理由写法："opposite angles of a cyclic quadrilateral sum to 180°" 或 "exterior angle of a cyclic quadrilateral equals the interior opposite angle"。
6. Theorem 5: The Alternate Segment Theorem | 定理五：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. If the tangent at A touches the circle and AB is a chord, then the angle between the tangent and AB equals the angle which AB subtends at any point on the arc on the other side of AB.
切线与弦在切点处所夹的角，等于该弦在另一侧弓形中所对的圆周角。若 A 处的切线接触圆，AB 是一条弦，则切线与 AB 的夹角等于 AB 在另一侧弧上任一点处所对的角。
Angle (tangent, AB) = angle in the alternate segment
t A and points C and D on the circle, if C lies in the segment opposite to angle TAB, then ∠TAB = ∠ACB. The word "alternate" he key: the angle you compare with must lie in the segment on the other side of the chord.
在标注好的图中，若 A 处的切线为 TA，圆上另有 C、D 两点，且 C 位于 ∠TAB 对面的弓形中，则 ∠TAB = ∠ACB。关键词是"alternate（另一侧）"：与之相等的角必须位于弦的另一侧弓形内。
There are two tangent-chord angles at each point of contact, one on each side of the chord, and they pair with two different angles in the two different segments. Students lose marks by pairing the wrong ones, so always draw the chord and shade the two segments before deciding.
每个切点处有两个弦切角，分别位于弦的两侧，它们对应两个不同弓形中的两个角。学生常因配错而失分，所以决定之前先把弦画出来，并把两个弓形分别涂色。
Typical use: a tangent and a chord are given, and you must find an angle at the circumference that no other theorem can reach. If a tangent appears in the diagram and a chord is drawn from the point of contact, the alternate segment theorem is almost always part of the solution.
典型用法：题目给出切线和弦，要求求一个其他定理都够不到的圆周角。若图中出现切线，且从切点画出了一条弦，那么弦切角定理几乎一定是解题的一部分。
7. Theorem 6: Tangents and Radii | 定理六：切线与半径
A tangent to a circle is perpendicular to the radius drawn to the point of contact. This is the definition-level fact behind every tangent calculation, and it creates a right angle that lets you use Pythagoras or trigonometry.
圆的切线垂直于过切点的半径。这是所有切线计算的底层事实，它制造出直角，从而让你可以使用勾股定理或三角函数。
TA ⊥ OA
The converse is also true and is used in proofs: if a line is perpendicular to a radius at a point on the circumference, that line is a tangent to the circle. Equivalently, the perpendicular distance from the centre to a tangent equals the radius.
逆定理同样成立且在证明中常用：若一条直线在圆上一点处垂直于半径，则这条直线是该圆的切线。等价地说，圆心到切线的垂直距离等于半径。
The line from T to the centre O bisects the angle between the tangents, and it also bisects the angle AOB between the two radii. Triangles OAT and OBT are congruent right-angled triangles, which is why the symmetry works.
从同一点 T 向圆引两条切线时，两条切线长度相等：TA = TB。连结 T 与圆心 O 的直线平分两条切线之间的夹角，同时平分两条半径之间的角 ∠AOB。三角形 OAT 与 OBT 是全等的直角三角形，这正是对称性成立的原因。
common exam shape: a quadrilateral OATB with two right angles at A and B. If you are told ∠ATB = 48°, then ∠ − 48° = 132°, and each base angle of the isosceles triangle TAB is (180° − 48°) ÷ 2 = 66°.
这产生了一个极常见的考试图形：四边形 OATB 在 A、B 两处各有一个直角。若已知 ∠ATB = 48°，则 ∠AOB = 360° − 90° − 90° − 48° = 132°，而等腰三角形 TAB 的两个底角各为 (180° − 48°) ÷ 2 = 66°。
Standard reason: "a tangent is perpendicular to the radius at the point of contact" or "tangents from an external point are equal in length".
规范理由写法："a tangent is perpendicular to the radius at the point of contact" 或 "tangents from an external point are equal in length"。
8. Theorem 7: Chords and the Perpendicular from the Centre | 定理七：垂径定理
ne also bisects the arc AB, and it bisects the angle at the centre standing on that chord.
从圆心向弦作垂线，垂足平分该弦。若 OM ⊥ AB 且 M 在 AB 上，则 AM = MB。同一条直线也平分弧 AB，并平分该弦所对的圆心角。
OM ⊥ AB → AM = MB
rpendicular distance d from the centre to the chord, and they satisfy the Pythagorean relation r² = d² + (half-chord)².
这条定理把弦的问题变成直角三角形的问题。涉及的三个长度是半径 r、半弦长，以及圆心到弦的垂直距离 d，它们满足勾股关系 r² = d² + (半弦长)²。
and d² = 10² − 8² = 100 − 64 = 36, giving d = 6 cm. Two steps, both earning method marks.
例题：半径 10 cm 的圆中有一条长 16 cm 的弦。圆心向弦作垂线平分该弦，故半弦长为 8 cm，于是 d² = 10² − 8² = 100 − 64 = 36，得 d = 6 cm。两步都能拿到方法分。
Two related facts appear in harder questions. Equal chords are equidistant from the centre, and the perpendicular bisector of any chord passes through the centre. The second fact is the usual way to find the centre of a circle given only an arc.
较难的题目中还会出现两个相关结论：等长的弦到圆心的距离相等；任意一条弦的垂直平分线都经过圆心。后一个结论正是只给一段弧时确定圆心位置的常用方法。
9. Worked Multi-Step Example | 多步综合例题
Multi-step questions almost never need only one theorem. The technique is to work outwards from the information you are given, writing the reason each time, until you reach the required angle. Numbers in a diagram that are not used at first usually become useful at step three or four.
多步综合题几乎不可能只用到一条定理。技巧是从已知条件出发逐步向外推进，每一步都写出理由，直到求出所需的角。图中一开始用不上的数字，往往在第三、四步就会派上用场。
Example A. A, B, C and D lie on a circle with centre O, in that order, and ∠ABC = 52°. Find ∠ADC, then find the reflex angle ∠AOC. Since ABCD is cyclic, ∠ADC = 180° − 52° = 128°. The reflex angle at the centre standing on arc ADC is twice ∠ABC, that is 2 × 128° = 256°.
例题 A：A、B、C、D 按顺序在圆心为 O 的圆上，且 ∠ABC = 52°。求 ∠ADC，再求优角 ∠AOC。因为 ABCD 为圆内接四边形，所以 ∠ADC = 180° − 52° = 128°。弧 ADC 所对的优角圆心角等于 ∠ABC 的两倍，即 2 × 128° = 256°。
Example B. TA and TB are tangents from an external point T, and ∠ATB = 48°. Triangles OAT and OBT are right-angled at A and B, so ∠AOB = 360° − 90° − 90° − 48° = 132°. Because TA = TB, triangle TAB is isosceles and ∠TAB = ∠ e alternate segment theorem, the angle in the segment on the other side of AB is also 66°.
例题 B：TA、TB 是从外点 T 引出的两条切线，∠ATB = 48°。三角形 OAT 与 OBT 在 A、B 处为直角，所以 ∠AOB = 360° − 90° − 90° − 48° = 132°。由于 TA = TB，三角形 TAB 为等腰三角形，∠TAB = ∠TBA = 66°。根据弦切角定理，AB 另一侧弓形中的圆周角也等于 66°。
Notice how each sentence in these examples ends with a reason. If you can produce four reasoned steps in a row, you are performing at the top of the mark scheme. If you can produce the right angle but cannot name the theorem, you are capped at roughly half marks.
注意这些例题中每一句都以理由结尾。能连续写出四步有理由的推导，就达到了评分标准的最高档；只会算出角却叫不出定理名，通常只能拿到一半左右的分数。
10. The Reasoning Language Examiners Want | 考官认可的规范表述
Reasoning marks are awarded for the correct named theorem, not for a vague gesture at the diagram. Learn each phrase as a fixed block of text. In an examination you should be able to write any of these in under five seconds.
理由分给的是正确的定理名称，而不是含糊的指图。请把每个表述当作固定短语来背。考场上要做到五秒内写出任意一条。
| style="border:1px solid
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 |