📚 Circle Theorems for IGCSE Mathematics | IGCSE 数学圆定理精讲
Circle theorems are a central part of the IGCSE Mathematics syllabus. They describe angle and length relationships within circles, and many exam questions require you to identify which theorem applies, give a reason, and calculate an unknown angle or length.
圆定理是 IGCSE 数学大纲中的核心内容。它们描述了圆内角和边长的关系，许多考试题目要求你识别适用的定理、给出理由并计算未知角或未知长度。
1. What Are Circle Theorems? | 什么是圆定理？
A circle theorem is a geometrical rule that links angles, chords, tangents, and segments of a circle. In IGCSE exams, you usually see diagrams with unknown angles labelled x, y, or z. You must select the correct theorem and state it as a reason, such as 'angle at centre is twice angle at circumference'.
圆定理是将圆的角、弦、切线和弓形联系起来的几何规则。在 IGCSE 考试中，你通常会看到带有未知角 x、y 或 z 的图形。你必须选择正确的定理并将其作为理由陈述，例如 '圆心角等于两倍圆周角'。
Before using circle theorems, you should be confident with basic angle facts: angles in a triangle sum to 180°, angles on a straight line sum to 180°, angles around a point sum to 360°, and base angles of an isosceles triangle are equal. These facts are often used together with circle theorems.
在使用圆定理之前，你应当熟悉基本的角关系：三角形内角和为 180°，直线上的角之和为 180°，绕一点一周的角之和为 360°，以及等腰三角形的底角相等。这些事实经常与圆定理一起使用。
2. The Angle at the Centre Theorem | 圆心角定理
If an arc AB subtends angle ∠AOB at the centre O and angle ∠APB at the circumference, then the central angle is twice the angle at the circumference. The point P must lie on the same side of chord AB as the centre.
如果弧 AB 在圆心 O 处张成角 ∠AOB，在圆周上的点 P 处张成角 ∠APB，则圆心角等于圆周角的两倍。点 P 必须与圆心位于弦 AB 的同一侧。
∠AOB = 2 × ∠APB
For example, if ∠APB = 38°, then ∠ ntral angle is known, divide by 2 to find the circumference angle. The standard reason is: 'angle at centre is twice angle at circumference'.
例如，如果 ∠APB = 38°，则 ∠AOB = 2 × 38° = 76°。反过来，如果已知圆心角，则除以 2 即可得到圆周角。标准的理由表述为：'圆心角等于两倍圆周角'。
3. The Angle in a Semicircle | 半圆角定理
n any angle subtended by AB at a point P on the circumference is a right angle. This means ∠APB = 90° for any point P on the circumference, as long as P is not A or B.
如果 AB 是圆的直径，则圆周上任意一点 P 对直径 AB 所张成的角都是直角。这意味着对于圆周上除 A、B 外的任意点 P，∠APB = 90°。
∠APB = 90°
This theorem is very useful when a diameter is given. It allows you to prove that a triangle is right-angled or to find a missing angle using the angle sum of a triangle. The standard reason is: 'angle in a semicircle is a right angle'.
当给出直径时，这个定理非常有用。它允许你证明一个三角形是直角三角形，或利用三角形内角和来求未知角。标准的理由表述为：'半圆上的圆周角是直角'。
4. Angles in the Same Segment | 同弓形角定理
rc in the same segment are equal. If points P and Q lie on the same side of chord AB, then ∠APB = ∠AQB. The chord AB creates two segments, and equality only holds within the same segment.
同一条弦或弧在同一弓形内所对的圆周角相等。如果点 P 和 Q 位于弦 AB 的同一侧，则 ∠APB = ∠AQB。弦 AB 会形成两个弓形，只有位于同一弓形内的圆周角才相等。
∠APB = ∠AQB
For example, if ∠ other angle subtended by chord AB in the same segment is also 47°. The standard reason is: 'angles in the same segment are equal'. A common error is applying this theorem to points on opposite sides of the chord.
例如，如果 ∠APB = 47°，则同一弓形内对弦 AB 所张成的任何其他圆周角也为 47°。标准的理由表述为：'同弓形上的圆周角相等'。一个常见错误是对位于弦两侧的点误用此定理。
5. Cyclic Quadrilaterals | 圆内接四边形定理
teral, opposite angles sum to 180°. If the vertices are labelled A, B, C, D in order, then ∠A + ∠C = 180° and ∠B + ∠D = 180°.
圆内接四边形是指四个顶点都在同一个圆周上的四边形。在任意圆内接四边形中，对角之和为 180°。如果顶点按顺序标记为 A、B、C、D，则 ∠A + ∠C = 180° 且 ∠B + ∠D = 180°。
∠A + ∠C = 180° and ∠B + ∠D = 180°
the exterior angle at any vertex is equal to the interior opposite angle. For example, if ∠A = 105°, then its opposite angle ∠C = 75°. The standard reason is: 'opposite angles in a cyclic quadrilateral sum to 180°'.
此外，任意顶点处的外角等于它的内对角。例如，如果 ∠A = 105°，则它的对角 ∠C = 75°。标准的理由表述为：'圆内接四边形对角互补'。
6. Tangent and Radius | 切线与半径定理
A tangent is a straight line that touches a circle at exactly one point. The radius drawn to the point of contact is perpendicular to the tangent. If O is the centre and T is the point of contact, then OT ⊥ PT, where PT is the tangent line.
切线是一条与圆恰好相切于一点的直线。连接圆心到切点的半径垂直于切线。如果 O 为圆心，T 为切点，则 OT ⊥ PT，其中 PT 为切线。
OT ⊥ PT ⇒ ∠OTP = 90°
This theorem is often used to create a right-angled triangle so that Pythagoras' theorem or trigonometry can be applied. The standard reason is: 'radius meets tangent at 90°'.
这个定理经常用于构造直角三角形，以便使用勾股定理或三角函数。标准的理由表述为：'半径与切线成 90°'。
7. Tangents from an External Point | 外点切线长定理
If two tangents are drawn from the same external point T to a circle, touching the circle at P and Q, then the two tangent segments TP and TQ are equal in length. Also, the line joining the external point to the centre O bisects the angle between the tangents.
如果从同一个外部点 T 引两条切线到圆上，切点分别为 P 和 Q，则两条切线段 TP 和 TQ 的长度相等。同时，连接外部点到圆心 O 的直线平分两条切线之间的角。
TP = TQ and ∠OTP = ∠OTQ
This theorem is useful when proving triangle congruence or finding unknown lengths. The standard reason is: 'tangents from a point to a circle are equal in length'.
这个定理在证明三角形全等或求未知长度时非常有用。标准的理由表述为：'从同一点引向圆的两条切线长度相等'。
8. The Alternate Segment Theorem | 弦切角定理
lternate segment. Suppose C is a point on the circumference in the alternate segment, then ∠BAT = ∠BCA, where T lies on the tangent.
如果过点 A 作一条切线和一条弦 AB，则切线与弦之间的角等于交替弓形中的角。假设 C 是交替弓形中圆周上的一点，则 ∠BAT = ∠BCA，其中 T 位于切线上。
∠BAT = ∠BCA
This theorem is often considered the hardest to spot in IGCSE questions because the tangent and chord form the angle outside the circle. The standard reason is: 'alternate segment theorem'.
在 IGCSE 题目中，弦切角定理通常被认为是最难识别的一个，因为切线和弦构成的角位于圆外。标准的理由表述为：'弦切角定理'。
9. Chords and Perpendicular Bisectors | 弦与垂直平分线定理
nversely, the perpendicular bisector of any chord passes through the centre of the circle.
从圆心 O 向弦 AB 所作的垂线平分该弦。如果 M 是圆心 O 到 AB 的垂足，则 AM = MB。反过来，任意弦的垂直平分线都经过圆心。
OM ⊥ AB ⇒ AM = MB
This theorem helps to find the distance from the centre to a chord, the length of a chord, or the radius when combined with Pythagoras' theorem. The standard reason is: 'perpendicular from centre to chord bisects chord'.
这个定理结合勾股定理，可以帮助求圆心到弦的距离、弦长或半径。标准的理由表述为：'圆心到弦的垂线平分弦'。
10. How to Apply Circle Theorems in IGCSE Exams | 如何在 IGCSE 考试中应用圆定理
In a typical IGCSE circle theorem question, you should first mark all given angles, radii, and tangents on the diagram. Next, look for isosceles triangles formed by two radii, because their base angles are equal. Then identify which circle theorem matches the given information.
在典型的 IGCSE 圆定理题目中，你应首先在图上标出所有已知角、半径和切线。接着，寻找由两条半径形成的等腰三角形，因为它们的底角相等。然后判断哪一个圆定理与已知信息相匹配。
Always write the reason for each calculation, as the mark scheme usually awards a mark for the theorem reason. Finally, use basic angle facts such as triangle angle sum, straight line angle sum, and angles around a point to finish the solution.
每一步计算都要写出理由，因为评分标准通常会给定理理由一个分数。最后，使用基本的角关系，如三角形内角和、直线角之和以及绕一点一周的角之和来完成解答。
- Step 1: Label the diagram.
第一步：标注图形。 - Step 2: Find isosceles triangles and right angles.
第二步：寻找等腰三角形和直角。 - Step 3: State the circle theorem used.
第三步：陈述所用圆定理。 - Step 4: Calculate unknown angles and lengths.
第四步：计算未知角和长度。
11. Common Mistakes and Proof Tips |
Published by TutorHao | IGCSE Mathematics Revision Series | aleveler.com
更多咨询请联系16621398022（同微信）
或截图后用「扫一扫 → 相册」选择本图