📚 IGCSE Mathematics: Intersecting Chord Theorem and Its Corollaries | IGCSE数学：相交弦定理及其推论
The Intersecting Chord Theorem is a key result in circle geometry, often tested in Edexcel IGCSE Mathematics. It connects the lengths of segments formed when two chords intersect inside a circle, and its corollaries extend this idea to secants and tangents. Understanding this theorem helps you solve a wide range of problems involving ratios, products, and unknown lengths.
相交弦定理是圆几何中的一个重要结论，在 Edexcel IGCSE 数学考试中经常出现。它揭示了两条弦在圆内相交时，所分得的各线段长度之间的数量关系；其推论进一步将该关系推广到割线与切线。掌握这一定理，有助于你解决大量涉及比例、乘积和未知长度的题目。
1. The Intersecting Chord Theorem | 相交弦定理的基本内容
If two chords AB and CD intersect at a point P inside a circle, then the product of the segments of one chord equals the product of the segments of the other chord. In symbols:
若圆内两条弦 AB 和 CD 相交于点 P，则一条弦被分成的两段长度之积等于另一条弦被分成的两段长度之积。用符号表示为：
AP × PB = CP × PD
This holds regardless of where the intersection point lies inside the circle, as long as both lines are chords of the same circle.
只要两条线段都是同一圆内的弦，无论交点位于圆内何处，该等式始终成立。
2. Why the Theorem Works | 定理的几何证明思路
To prove the theorem, draw chords AC and BD. Consider triangles APC and DPB. Because angles subtended by the same arc are equal, angle ACP equals angle DBP and angle CAP equals angle BDP. Therefore the two triangles are similar.
为了证明这一定理，可连接 AC 和 BD，考察三角形 APC 与三角形 DPB。由于同弧所对的圆周角相等，可得 ∠ACP = ∠DBP，∠CAP = ∠BDP。因此这两个三角形相似。
From similar triangles, the ratios of corresponding sides are equal:
由相似三角形的对应边成比例，可得：
AP / CP = DP / PB
Cross-multiplying gives AP × PB = CP × PD, which proves the theorem.
交叉相乘后得到 AP × PB = CP × PD，定理得证。
3. Worked Example 1: Finding an Unknown Length | 例题 1：求未知线段长度
Two chords AB and CD intersect at point P inside a circle. It is given that AP = 4 cm, PB = 6 cm, and CP = 3 cm. Find the length of PD.
圆内两弦 AB 和 CD 相交于点 P。已知 AP = 4 cm，PB = 6 cm，CP = 3 cm，求 PD 的长度。
Using the Intersecting Chord Theorem:
应用相交弦定理：
AP × PB = CP × PD
4 × 6 = 3 × PD
24 = 3 × PD
PD = 8 cm
So the unknown segment PD has length 8 cm.
因此，未知线段 PD 的长度为 8 cm。
4. The Secant-Secant Power Theorem | 割线定理（推论一）
If two secants PAB and PCD are drawn from an external point P to a circle, intersecting the circle at points A, B and C, D respectively, then:
若从圆外一点 P 引两条割线 PAB 和 PCD，分别与圆交于点 A、B 和 C、D，则有：
PA × PB = PC × PD
Here PA is the shorter external segment and PB is the full secant length from P to the farther intersection point B.
其中 PA 是较短的圆外部分，PB 是从 P 到较远交点 B 的完整割线长度。
This is a direct corollary of the Intersecting Chord Theorem when the intersection point is moved outside the circle.
这是相交弦定理中交点移到圆外时的直接推论。
5. Worked Example 2: Secant Theorem in Action | 例题 2：割线定理的应用
A point P lies outside a circle. Two secants from P meet the circle as follows: PA = 5 cm, AB = 7 cm, and PC = 6 cm. Find PD.
圆外一点 P，从 P 引两条割线与圆相交：PA = 5 cm，AB = 7 cm，PC = 6 cm，求 PD。
First note that PB = PA + AB = 5 + 7 = 12 cm. Applying the Secant Theorem:
首先注意 PB = PA + AB = 5 + 7 = 12 cm。应用割线定理：
PA × PB = PC × PD
5 × 12 = 6 × PD
60 = 6 × PD
PD = 10 cm
Therefore the full secant length PD is 10 cm.
因此，完整的割线长度 PD 为 10 cm。
6. The Tangent-Secant Power Theorem | 切割线定理（推论二）
If a tangent from an external point P touches the circle at point T, and a secant from P intersects the circle at points A and B, then:
若从圆外一点 P 引圆的切线，切点为 T，同时从 P 引一条割线与圆交于点 A 和 B，则有：
PT² = PA × PB
This is also called the Tangent-Secant Theorem. It can be viewed as a limiting case of the secant theorem where the two intersection points C and D coincide at T.
这被称为切割线定理。它可以看作是割线定理中两个交点 C 和 D 重合于点 T 时的极限情形。
Note that PA is the shorter segment from P to the nearer intersection point A, and PB is the full secant length.
注意 PA 是从 P 到较近交点 A 的较短线段，PB 是完整割线长度。
7. Worked Example 3: Tangent-Secant Theorem | 例题 3：切割线定理应用
A tangent from point P touches a circle at T. A secant from P intersects the circle at A and B, with PA = 4 cm and AB = 12 cm. Find the tangent length PT.
从点 P 引圆的切线，切点为 T。从 P 引一条割线交圆于 A 和 B，其中 PA = 4 cm，AB = 12 cm，求切线 PT 的长度。
First compute PB = PA + AB = 4 + 12 = 16 cm. Applying the Tangent-Secant Theorem:
先计算 PB = PA + AB = 4 + 12 = 16 cm。应用切割线定理：
PT² = PA × PB
PT² = 4 × 16 = 64
PT = √64 = 8 cm
Therefore the tangent length is 8 cm.
因此，切线长度为 8 cm。
8. Chord Bisection and Perpendicularity | 垂径定理与弦平分
Another important corollary is that if a line from the centre of a circle is perpendicular to a chord, then it bisects the chord. Conversely, if a line from the centre bisects a chord, it is perpendicular to the chord.
另一个重要推论是：从圆心向弦作垂线，则垂线平分该弦；反之，从圆心连接弦中点的直线必垂直于该弦。
This follows from congruent triangles formed by the two radii and the chord. Since both radii are equal, the perpendicular from the centre creates two congruent right triangles.
这是由半径相等形成的全等直角三角形推导出来的。由于两条半径相等，圆心向弦作垂线会形成两个全等的直角三角形。
When solving problems, if you know the radius and the distance from the centre to the chord, you can find half the chord length using Pythagoras’ Theorem.
解题时，若已知半径和弦心距，可利用勾股定理求出弦长的一半。
9. Worked Example 4: Radius and Chord Distance | 例题 4：半径与弦心距
A circle has radius 10 cm. The perpendicular distance from the centre to a chord is 6 cm. Find the length of the chord.
已知圆的半径为 10 cm，圆心到弦的垂直距离为 6 cm，求弦长。
Let the chord be AB and let M be its midpoint. Then OM = 6 cm and OA = 10 cm, where O is the centre. Triangle OMA is right-angled at M, so:
设弦为 AB，M 为其中点。则 OM = 6 cm，OA = 10 cm，其中 O 为圆心。三角形 OMA 在 M 处为直角三角形，因此：
AM² = OA² − OM² = 10² − 6² = 100 − 36 = 64
AM = 8 cm
Therefore the full chord AB = 2 × 8 = 16 cm.
因此，弦 AB 的全长为 2 × 8 = 16 cm。
10. Intersecting Chords with Algebraic Expressions | 含代数表达式的相交弦问题
In IGCSE examinations, lengths are often given as algebraic expressions. For example, two chords intersect and one segment is x, another is x + 2, a third is 3, and the fourth is 8. Using the theorem:
在 IGCSE 考试中，线段长度常以代数表达式表示。例如，两条弦相交，其中一段为 x，另一段为 x + 2，第三段为 3，第四段为 8。应用定理：
x(x + 2) = 3 × 8 = 24
x² + 2x − 24 = 0
(x + 6)(x − 4) = 0
Since x must be positive, x = 4. Always discard the negative root in length problems.
由于长度必须为正数，所以 x = 4。在长度问题中应舍去负根。
11. Common Mistakes and Exam Tips | 常见错误与考试建议
- Use the correct segments: For the intersecting chord theorem, use the products of the two parts of each chord, not the full chord lengths.
- 注意线段选取：运用相交弦定理时，应使用每条弦被交点分成的两段之积，而不是整条弦的长度。
- Distinguish internal vs external: The chord theorem applies inside the circle; the secant and tangent theorems apply from an external point.
- 区分圆内与圆外：相交弦定理适用于圆内交点；割线定理和切割线定理适用于圆外一点。
- Check units: All lengths must be in the same unit before applying any formula.
- 检查单位：应用公式前，所有长度必须保持同一单位。
- Draw a clear diagram: Label every known length and mark the point of intersection.
- 画清晰图形：标出所有已知长度，并标明交点位置。
- Pythagoras with chords: When the distance from the centre to a chord is involved, remember to use the right triangle formed by the radius, half the chord, and the perpendicular distance.
- 弦与勾股定理：涉及弦心距时，利用半径、半弦长和弦心距组成的直角三角形。
12. Practice Problems | 巩固练习
Try these problems on your own before checking the answers.
请独立完成以下练习，再对照答案。
- Problem 1: Two chords AB and CD intersect at P. AP = 3 cm, PB = 8 cm, CP = 4 cm. Find PD.
- 题目 1：两弦 AB 与 CD 相交于 P。AP = 3 cm，PB = 8 cm，CP = 4 cm，求 PD。
- Problem 2: From an external point P, a tangent PT has length 6 cm, and a secant has external segment PA = 4 cm. Find the full length PB.
- 题目 2：从圆外一点 P，切线 PT = 6 cm，割线的外部线段 PA = 4 cm，求完整割线长度 PB。
- Problem 3: A circle of radius 13 cm has a chord 24 cm long. Find the perpendicular distance from the centre to the chord.
- 题目 3：半径为 13 cm 的圆内有一条长为 24 cm 的弦，求圆心到弦的垂直距离。
Answers: 1) PD = 6 cm 2) PB = 9 cm 3) distance = 5 cm
Solutions: For Problem 1, 3 × 8 = 4 × PD, so PD = 6. For Problem 2, PT² = PA × PB, so 36 = 4 × PB, hence PB = 9. For Problem 3, half the chord is 12 cm, so d = √(13² − 12²) = √25 = 5 cm.
解答：第 1 题，3 × 8 = 4 × PD，所以 PD = 6。第 2 题，PT² = PA × PB，即 36 = 4 × PB，所以 PB = 9。第 3 题，半弦长为 12 cm，则弦心距 d = √(13² − 12²) = √25 = 5 cm。
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