📚 Intersecting Chords Theorems | 相交弦定理
In IGCSE circle geometry, three closely related theorems describe the relationship between the segments formed when chords, secants, or a tangent and a secant intersect. Known collectively as the Intersecting Chords Theorems, these results are powerful tools for calculating unknown lengths inside and outside a circle. Understanding their statements, proofs, and typical applications will prepare you for the most common Edexcel exam questions on this topic.
在IGCSE圆的几何中，有三个密切相关的定理描述了弦、割线或切线与割线相交时形成的线段之间的关系。这些定理统称为相交弦定理，是计算圆内外未知长度的有力工具。理解它们的陈述、证明和典型应用，将帮助你应对爱德思考试中关于这一主题最常见的题目。
1. Introduction to Circle Theorems and Chords | 圆定理与弦简介
A chord is a line segment whose endpoints both lie on the circle. The diameter is the longest chord. When two chords intersect, they cut each other into smaller pieces called segments of the chord. The Intersecting Chords Theorem tells us that the products of the lengths of the segments of each chord are equal.
弦是两端点都在圆上的线段。直径是最长的弦。当两条弦相交时，它们彼此分割成更小的线段，称为弦的截段。相交弦定理告诉我们，每条弦被交点分成的两段长度的乘积相等。
There are three variants of the theorem you need to know for Edexcel IGCSE:
对于爱德思IGCSE，你需要掌握定理的三种形式：
- Internal intersection – two chords meet inside the circle.
- External intersection – two secants (extended chords) meet outside the circle.
- Tangent and secant – a tangent and a secant meet outside the circle.
- 内相交 – 两弦在圆内相交。
- 外相交 – 两条割线（弦的延长线）在圆外相交。
- 切线与割线 – 切线与割线在圆外相交。
These rules are sometimes grouped under the heading "Power of a Point" theorems, but in IGCSE they appear as separate, memorisable results.
这些规则有时被归类为“点幂定理”，但在IGCSE中它们作为独立可记的结果出现。
2. Intersecting Chords Theorem (Inside the Circle) | 相交弦定理（圆内相交）
If two chords AB and CD of a circle intersect at a point P inside the circle, then the products of the lengths of the segments of each chord are equal:
若圆的两条弦AB与CD在圆内一点P相交，则每条弦被交点分成的两段长度的乘积相等：
AP × PB = CP × PD
In words: the product of the segments of one chord equals the product of the segments of the other. This is true regardless of where the chords cross, as long as the point of intersection is inside the circle.
用文字表述：一条弦两段长的乘积等于另一条弦两段长的乘积。无论弦在何处相交，只要交点在圆内，该等式都成立。
It is essential to label the segments correctly. If a chord PQ is cut by another chord at point X, then the two segments are PX and XQ. Always multiply the two parts of the same chord together, not across different chords.
正确标记截段至关重要。若弦PQ被另一条弦在X点分割，则两段分别为PX和XQ。始终将同一条弦的两部分相乘，而不是在不同弦之间交叉相乘。
3. Proof of the Intersecting Chords Theorem | 相交弦定理的证明
For Edexcel IGCSE, you are not required to reproduce the proof, but understanding one version deepens geometric reasoning. The standard proof uses similar triangles formed by joining A to D and C to B.
爱德思IGCSE不要求你写出证明过程，但理解一种证法可以加深几何推理。标准证明利用连接A与D、C与B后形成的相似三角形。
In the diagram, angle ∠DAP equals angle ∠BCP because they subtend the same arc BD. Similarly, angle ∠ADP equals angle ∠CBP because they subtend arc AC. Thus, triangle APD is similar to triangle CPB (AA criterion).
在图中，∠DAP等于∠BCP，因为它们对着同一条弧BD。同样，∠ADP等于∠CBP，因为它们对着弧AC。因此，三角形APD相似于三角形CPB（AA判定）。
From the similarity: AP/CP = PD/PB. Cross-multiplying gives AP × PB = CP × PD.
由相似关系：AP/CP = PD/PB。交叉相乘即得AP × PB = CP × PD。
This proof holds when the intersection is inside. When the point is outside, similar ideas with secants lead to the external forms.
当交点在圆内时此证明成立。当点在圆外时，利用割线的类似思路可导出外部形式。
4. Applying the Theorem to Find Missing Lengths | 应用定理求未知长度
chords AB and CD intersect at point P. AP = 3 cm, PB = 4 cm, CP = x cm, PD = 6 cm. Find x.
例题：圆中，弦AB与CD交于点P。AP = 3 cm，PB = 4 cm，CP = x cm，PD = 6 cm。求x。
Using AP × PB = CP × PD → 3 × 4 = x × 6 → 12 = 6x → x = 2 cm.
利用AP × PB = CP × PD → 3 × 4 = x × 6 → 12 = 6x → x = 2 cm。
Another typical problem gives three segment lengths and asks for the fourth. Always set up the equation so that the unknown is isolated. Check that your answer makes physical sense – a segment length cannot be negative, and it must be smaller than the diameter if inside the circle.
另一种典型题目给出三个截段长度，求第四个。建立方程时始终让未知数独立。检查答案的物理意义——截段长度不能为负，如果在圆内，它必须小于直径。
s a point on AB such that AP = 4 cm, then PB = 6 cm. Watch for these partial-length clues.
有时图会显示整条弦的长度和一个截段。例如，若AB = 10 cm，P是AB上一点且AP = 4 cm，则PB = 6 cm。注意这些部分长度的线索。
5. Intersecting Secants Theorem (Outside the Circle) | 割线相交定理（圆外相交）
A secant is a line that cuts a circle at two points. If two secants PAB and PCD are drawn from an external point P, meeting the circle at A, B and C, D respectively (with P–A–B and P–C–D ordering, so A and C are nearer to P), then:
割线是一条与圆相交于两点的直线。若从圆外一点P引两条割线PAB和PCD，分别交圆于A、B和C、D（顺序为P–A–B和P–C–D，即A和C离P较近），则：
PA × PB = PC × PD
Notice the similarity with the internal version. The entire secant segment from the external point to the far intersection (PB or PD) multiplied by its external part (PA or PC) is constant.
注意它与内相交形式的相似性。从圆外点到较远交点的整条割线段（PB或PD）乘以其外部部分（PA或PC）为定值。
The proof again relies on similar triangles: connect B to C and A to D, then show triangle PAD is similar to triangle PCB. The key is that ∠P is common and angles ∠PAD and ∠PCB subtend the same arc BD.
证明同样依赖相似三角形：连接B与C、A与D，然后证明三角形PAD相似于三角形PCB。关键是∠P为公共角，且∠PAD与∠PCB对着同一条弧BD。
In IGCSE problems, you will typically be given three of the four lengths PA, PB, PC, PD and asked to find the fourth. Always write down the full equation before substituting.
IGCSE题目中，通常给出PA、PB、PC、PD四个长度中的三个，求第四个。始终在代入之前完整写出等式。
6. Tangent-Secant Theorem | 切线-割线定理
The third important case occurs when one of the lines from point P is a tangent (touching the circle at point T) and the other is a secant PAB (intersecting at A and B, with A nearer to P). Then:
第三种重要情况是，从点P引出的线中一条是切线（切圆于点T），另一条是割线PAB（交圆于A和B，A离P较近）。此时：
PT² = PA × PB
Squaring the tangent length gives the product of the secant's whole length and its external segment. This is a special case of the secant theorem if we imagine the two secant intersections coalescing into the single point T.
切线长度的平方等于割线全长与其外部线段的乘积。如果我们将两条割线的两个交点视为合并为单一点T，这就是割线定理的特例。
The proof uses triangle PTA and PTB. Since PT is a tangent, ∠PTA equals the angle in the alternate segment (∠TBA). Then angle ∠ so triangle PTA is similar to triangle PTB. Hence PT/PA = PB/PT, leading to PT² = PA × PB.
证明利用三角形PTA和PTB。因为PT是切线，∠PTA等于弦切角（∠TBA）。又∠P为公共角，故三角形PTA相似于三角形PTB。从而PT/PA = PB/PT，推出PT² = PA × PB。
This theorem is tested frequently, often asking for the length of the tangent given the secant segments, or vice versa.
该定理考查频繁，常给出割线段求切线长度，或反之。
7. Worked Examples | 典型例题
Example 1 (Internal intersection): Chords XY and ZW intersect at P. XP = 5 cm, PY = 8 cm, ZP = 4 cm. Find PW.
例题1（内相交）：弦XY与ZW交于P。XP = 5 cm，PY = 8 cm，ZP = 4 cm。求PW。
Solution: XP × PY = ZP × PW → 5 × 8 = 4 × PW → 40 = 4 × PW → PW = 10 cm.
解：XP × PY = ZP × PW → 5 × 8 = 4 × PW → 40 = 4 × PW → PW = 10 cm。
Example 2 (External secants): e at A and B (PA = 6 cm, AB = 9 cm). Another secant meets at C and D (PC = 7 cm). Find CD.
例题2（外割线）：从圆外一点P，一条割线交圆于A和B（PA = 6 cm，AB = 9 cm）。另一条割线交圆于C和D（PC = 7 cm）。求CD。
tes PA × PB = PC × PD → 6 × 15 = 7 × PD → 90 = 7 × PD → PD = 90/7 ≈ 12.86 cm. Then CD = PD – PC = 12.86 – 7 = 5.86 cm (or exact: 90/7 – 7 = (90 – 49)/7 = 41/7 cm).
首先，PB = PA + AB = 6 + 9 = 15 cm。定理给出PA × PB = PC × PD → 6 × 15 = 7 × PD → 90 = 7 × PD → PD = 90/7 ≈ 12.86 cm。然后CD = PD – PC = 12.86 – 7 = 5.86 cm（或精确值：90/7 – 7 = 41/7 cm）。
Example 3 (Tangent-secant): at T. A secant from P meets the circle at A and B, with PA = 4 cm and AB = 5 cm. Find PT.
例题3（切线-割线）：从P引切线切圆于T，引割线交圆于A和B，PA = 4 cm，AB = 5 cm。求PT。
PB = PA + AB = 9 cm. PT² = PA × PB = 4 × 9 = 36 → PT = 6 cm.
PB = PA + AB = 9 cm。PT² = PA × PB = 4 × 9 = 36 → PT = 6 cm。
8. Common Misconceptions to Avoid | 常见误区避免
Mixing up segments of different chords: Always multiply the two pieces belonging to the same chord: AP × PB, not AP × CP.
混淆不同弦的截段：始终将同一条弦的两段相乘：AP × PB，而不是AP × CP。
Forgetting to add segments for the whole secant: In external cases, PB is the whole length from P to the far intersection, not just the part inside the circle.
忘记将截段相加得到整条割线：外相交情形中，PB是从P到较远交点的整段长度，而不只是在圆内的部分。
Using the tangent length unsquared: Remember PT² in the tangent-secant rule; a common error is writing PT = PA × PB.
切线长度未平方：记住切线-割线法则中PT²；常见错误是写成PT = PA × PB。
Applying internal rule to external points: The internal theorem only works when the intersection is strictly inside the circle.
将内相交规则用于外点：内相交定理仅在交点严格在圆内时成立。
Assuming chords must be diameters: The theorems hold for any two chords, regardless of their angle or length.
假设弦必须是直径：这些定理适用于任意两条弦，无论夹角或长短。
9. Diagram Interpretation and Labeling | 图形解读与标记
cut the circle twice, use PA × PB = PC × PD. If one line is a tangent, use PT² = PA × PB.
爱德思IGCSE的图形通常会展示圆、相交线以及标注点。你必须能根据交点的位置判断用哪条定理。若交点在圆内，用AP × PB = CP × PD。若在圆外且两条线都与圆交于两点，用PA × PB = PC × PD。若其中一条是切线，用PT² = PA × PB。
Practise redrawing the diagram yourself and colour-coding the segments you multiply. This reduces the risk of misreading which lengths correspond to which parts of the formula.
练习自己重绘图形，并对要相乘的截段进行颜色标记。这可降低误读长度对应关系而用错公式的风险。
uation using the appropriate theorem and solve for x. Keep all solutions positive and less than the diameter.
有时题目会给出代数式而非数字（例如AP = x，PB = x+3）。用适当定理建立方程，然后求解x。所有解应为正数且小于直径。
10. Practice Questions and Exam Tips | 练习题与应试技巧
Try these before moving on:
在继续之前试做以下题目：
- Two chords intersect inside a circle. Segments of one chord are 6 cm and x cm. Segments of the other are 8 cm and 9 cm. Find x. (Answer: 12 cm)
- 从圆外一点P，割线PAB满足PA = 5 cm, AB = 7 cm。同一点另一割线PCD满足PC = 4 cm。求CD。(答案：11 cm)
- A tangent PT = 8 cm, and a secant PAB has PA = 4 cm. Find PB. (Answer: 16 cm)
In the exam, always show your substitution step clearly. Even if you make an arithmetic error, you can earn method marks. State the theorem you are using: "By the intersecting chords theorem, ..."
考试时，务必清晰地写出代入步骤。即使计算有误，也能得到方法分。陈述你使用的定理：“由相交弦定理得……”。
Check your units – all lengths should be in the same unit. If a length turns out to be a fraction, leave it as an exact fraction unless the question specifies otherwise.
检查单位——所有长度应使用相同单位。如果长度是分数，保留精确分数，除非题目另有要求。
11. Linking to Other Circle Theorems | 与其他圆定理的联系
The intersecting chords theorems complement the wider family of circle theorems. For instance, the alternate segment theorem is used in the proof of the tangent-secant case. Angle at the centre, angles in the same segment, and cyclic quadrilateral properties frequently appear in the same IGCSE question alongside segment products.
相交弦定理与其他圆定理相辅相成。例如，切线-割线情形的证明用到了弦切角定理。圆心角定理、同弧上的圆周角以及圆内接四边形的性质常与线段乘积出现在同一IGCSE试题中。
When solving a multi-step problem, first identify whether you need angle facts to find a missing length, or length facts to find a missing angle. The intersecting chords theorems are pure length relationships, independent of angles, so they can often be applied directly once the correct segments are identified.
解决多步问题时，先判断是需要用角度关系求未知长度，还是用长度关系求未知角度。相交弦定理是纯粹的长度关系，与角度无关，因此一旦识别出正确的截段，通常可直接应用。
12. Summary | 总结
| Intersection type 相交类型 | Theorem 定理 | Example usage 示例用法 |
|---|---|---|
| Inside circle 圆内 | AP × PB = CP × PD | 3 × 4 = x × 6 |
| Outside (two secants) 圆外（两割线） | PA × PB = PC × PD | 5 × 12 = 4 × (4+CD) |
| Outside (tangent-secant) 圆外（切线与割线） | PT² = PA × PB | 8² = 4 × PB |
Memorise these three formulas, practise recognising the configuration from a diagram, and always double-check which lengths correspond to which parts of the product. With these tools, intersecting chords questions become straightforward marks on your IGCSE Edexcel Mathematics paper.
记住这三个公式，练习从图形中识别相应结构，并始终反复核对哪些长度对应乘积中的哪些部分。掌握这些工具后，相交弦题目将成为你IGCSE爱德思数学试卷上的简单得分项。
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