📚 Circle Theorems 2: Tangents, Chords and Secants | 圆定理2：切线、弦与割线
Building on the basic circle theorems - the angle at the centre is twice the angle at the circumference, the angle in a semicircle is 90°, angles in the same segment are equal, and opposite angles of a cyclic quadrilateral sum to 180° - this second part explores the theorems that involve tangents, chords and secants. These results turn a circle into a machine for generating right angles, equal lengths and similar triangles, and they are examined every year in Edexcel IGCSE Mathematics, almost always inside multi-step geometry problems rather than in isolation.
在基本圆定理（圆心角是圆周角的两倍、半圆上的圆周角为 90°、同弧所对的圆周角相等、圆内接四边形对角互补）的基础上，本文第二部分探讨涉及切线、弦和割线的定理。这些结论把圆变成一个能生成直角、等长线段和相似三角形的工具，并且每年都会在 Edexcel IGCSE 数学考试中出现，而且几乎总是出现在多步骤几何题中，而非单独考查。
1. Tangent-Radius Perpendicularity | 切线与半径垂直
A tangent to a circle is a straight line that touches the circle at exactly one point, called the point of contact. The most fundamental tangent theorem states that the radius drawn to the point of contact is perpendicular to the tangent. This single right angle is the reason the tangent theorems sit so naturally alongside Pythagoras' theorem and trigonometry.
圆的切线是一条与圆恰好只有一个公共点的直线，这个公共点称为切点。最基本的切线定理指出：连接切点的半径垂直于该切线。正是这个直角，使得切线定理能够与勾股定理、三角函数自然地结合使用。
In diagrammatic terms, if a tangent touches the circle at point T and O is the centre, then OT ⊥ tangent line at T. Because OT is a radius and the tangent is at 90° to it, triangle OPT (where P is any point on the tangent) is right-angled at T, so we can apply Pythagoras:
在图形中，如果切线在点 T 处与圆相切，O 为圆心，那么 OT ⊥ 切线（在 T 点处）。由于 OT 是半径且与切线成 90°，三角形 OPT（P 为切线上任意一点）在 T 处为直角，因此可以使用勾股定理：
OP² = OT² + PT²
= 12 cm. Notice that the radius and the tangent length form the two shorter sides, while the line from the centre to the external point is the hypotenuse.
例题：从 P 点引出的切线在 T 处与圆相切。已知 OT = 5 cm，OP = 13 cm，求 PT。代入得 PT² = 13² − 5² = 169 − 25 = 144，所以 PT = √144 = 12 cm。注意半径与切线长构成两条直角边，而从圆心到外点的连线是斜边。
2. Two Tangents from an External Point | 从外点引出的两条切线
y touch the circle at A and B, then PA = PB. Triangles OAP and OBP are congruent by RHS (OA = OB as radii, OP is common, and both have a right angle at A and B).
从圆外一点 P 可以画出恰好两条切线。第二条重要定理指出：这两条切线长度相等。若它们分别在 A、B 处与圆相切，则 PA = PB。三角形 OAP 与 OBP 依据 RHS 全等（OA = OB 是同圆半径，OP 为公共边，且在 A、B 处均为直角）。
nces follow from the same congruence. First, OP bisects the angle between the tangents, so ∠APO = ∠BPO. Second, OP also bisects the angle at the centre, so ∠AOP = ∠BOP. The quadrilateral OAPB is a kite with two right angles, which means ∠AOB + ∠APB = 180°, a fact that is frequently the key to a final angle in an exam question.
同一组全等三角形还带来两个推论。第一，OP 平分两切线之间的夹角，即 ∠APO = ∠BPO。第二，OP 也平分圆心角，即 ∠AOP = ∠BOP。四边形 OAPB 是一个含有两个直角的筝形，因此 ∠AOB + ∠APB = 180°，这一点常常是考试题中求出最后一个角的关键。
orked example: PA and PB are tangents, PA = 8 cm and OP = 17 cm. Find the radius. Since OA ⊥ AP is right-angled at A, so OA² = 17² − 8² = 289 − 64 = 225, giving OA = 15 cm. If instead ∠APB = 60° is given, then ∠AOB = 120° by the kite property, and each base angle ∠APO = 30°.
例题：PA 与 PB 是切线，PA = 8 cm，OP = 17 cm，求半径。因为 OA ⊥ PA，三角形 OAP 在 A 处为直角，故 OA² = 17² − 8² = 289 − 64 = 225，得 OA = 15 cm。若已知 ∠APB = 60°，由筝形性质可得 ∠AOB = 120°，且每个底角 ∠APO = 30°。
3. The Alternate Segment Theorem | 切弦角定理（交错弧定理）
The alternate segment theorem is the most elegant tangent result and the one students lose marks on most often. It states that the angle between a tangent and a chord drawn from the point of contact is equal to the angle in the alternate segment, that is, the angle subtended by the same chord at any point on the circumference on the other side of the chord.
切弦角定理（交错弧定理）是最优美的切线结论，也是学生最容易失分的一个。它指出：切线与从切点引出的弦之间的夹角，等于该弦在交错弧中所对的圆周角，也就是弦的另一侧圆周上任意一点对该弦所张的角。
rd TA make an angle of θ, and B is any point on the circle on the opposite side of TA from the tangent, then ∠ awn in opposite directions at T, there are always two such angles, and each equals the angle in its own alternate segment. A useful check is that the two tangent-chord angles at T sum to 180°.
用符号表示：若 T 处的切线与弦 TA 的夹角为 θ，B 是圆周上位于 TA 另一侧（与切线相对一侧）的任一点，则 ∠TBA = θ。由于在 T 处可以向两个相反方向画出切线段，因此总有两个这样的角，各等于其对应交错弧中的圆周角。一个有用的检验是：T 处两个切弦角之和为 180°。
tangent at T and chord TA make an angle of 55°, and B lies on the alternate segment. Then ∠BA = 55°. If the question also tells you that TA = TB, then triangle TAB is isosceles with ∠TAB = ∠TBA = 55°, leaving ∠ATB = 180° − 110° = 70°. Always quote the theorem by name when asked for a reason: "alternate segment theorem".
例题：T 处的切线与弦 TA 成 55° 角，B 位于交错弧上，则 ∠TBA = 55°。若题目还给出 TA = TB，则三角形 TAB 为等腰三角形，∠TAB = ∠TBA = 55°，于是 ∠ATB = 180° − 110° = 70°。要求说明理由时务必写出定理名称："切弦角定理（交错弧定理）"。
4. Perpendicular from the Centre Bisects a Chord | 垂径定理
pendicular drawn from the centre of a circle to a chord bisects that chord. In other words, if OM ⊥ o true: the line from the centre to the midpoint of a chord is perpendicular to the chord.
弦是连接圆周上两点的线段。从圆心向弦所作的垂线平分该弦。也就是说，若 OM ⊥ AB，其中 M 在 AB 上、O 为圆心，则 AM = MB。逆命题同样成立：从圆心连向弦中点的直线垂直于该弦。
This theorem creates a right-angled triangle whose hypotenuse is the radius. If the radius is r, the chord is 2d long and the distance from the centre to the chord is h, then the half-chord, distance and radius satisfy:
该定理构造出一个以半径为斜边的直角三角形。若半径为 r，弦长为 c，圆心到弦的距离为 h，则半弦长、距离与半径满足：
r² = h² + (c ÷ 2)²
ce is that the perpendicular bisector of any chord passes through the centre, which is exactly how the centre of a circle is located from three points on the circumference in construction questions.
例题：半径为 13 cm 的圆中有一条长 24 cm 的弦。半弦长为 12 cm，故 h² = 13² − 12² = 169 − 144 = 25，h = 5 cm。一个相关推论是：任意弦的垂直平分线都经过圆心，这正是作图题中由圆周上三点确定圆心的方法。
5. Equal Chords and Equal Distances from the Centre | 等弦与圆心等距
Equal chords in the same circle (or in congruent circles) are equidistant from the centre, and conversely, chords that are equidistant from the centre are equal in length. This follows directly from the previous theorem, since two chords at the same perpendicular distance h from a centre of radius r both have length 2√r² − h².
同一个圆（或等圆）中，相等的弦到圆心的距离相等；反之，到圆心距离相等的弦长度相等。这可由上一定理直接推出：两条弦若到圆心 r 的垂距同为 h，则长度都是 2√r² − h²。
A second useful result concerns the perpendicular bisector: the perpendicular bisector of a chord always passes through the centre of the circle. Combined with the previous section, this means that if you can construct two perpendicular bisectors of two chords, their intersection is the centre. This is the standard construction for finding a circle's centre and also proves that any three non-collinear points determine exactly one circle.
第二个有用的结论涉及垂直平分线：弦的垂直平分线必定经过圆心。与上一节结合可知，若作出两条弦的垂直平分线，其交点即为圆心。这是求圆心位置的标准作图法，同时也证明：任意三个不共线的点确定唯一一个圆。
Worked example: two parallel chords of a circle of radius 10 cm are 8 cm and 8 cm long. Each has half-length 4 cm, so each lies at distance √100 − 16 = √84 ≈ 18.3 cm if they lie on opposite sides of the centre, or zero offset if on the same side.
例题：半径 10 cm 的圆中有两条长 8 cm 的平行弦。每条弦的半弦长为 4 cm，故各自到圆心的距离为 √100 − 16 = √84 ≈ 9.17 cm。由于两弦相等，它们到圆心等距；若分居圆心两侧，则两弦间距为 2 × 9.17 ≈ 18.3 cm。
6. Intersecting Chords Theorem | 相交弦定理
When two chords AB and CD intersect at a point P inside the circle, the products of the two segments of each chord are equal:
当两条弦 AB 与 CD 在圆内一点 P 处相交时，各弦被分成的两段长度之积相等：
AP × PB = CP × PD
uses similar triangles. Joining A to C and B to D creates triangles APC and DPB, in which ∠APC = ∠DPB (vertically opposite) and ∠PAC = ∠ herefore similar, and equating the ratios of corresponding sides gives the product result.
证明利用相似三角形。连接 A 与 C、B 与 D，得到三角形 APC 与 DPB，其中 ∠APC = ∠DPB（对顶角），∠PAC = ∠PDB（同弧 BC 所对的圆周角）。因此两三角形相似，对应边成比例，即得乘积关系。
exam may give the total chord length rather than both segments, so subtract carefully. If AB = 10 cm and AP = 4 cm, then PB = 6 cm.
例题：弦 AB 与 CD 交于 P，AP = 4 cm，PB = 6 cm，CP = 3 cm，则 4 × 6 = 3 × PD，所以 PD = 24 ÷ 3 = 8 cm。该定理常被"伪装"：考题可能给出整条弦的长度而不是两段长度，因此要仔细相减。若 AB = 10 cm、AP = 4 cm，则 PB = 6 cm。
7. Intersecting Secants Theorem | 割线定理
A secant is a line that cuts the circle at two points. When two secants are drawn from the same external point P, meeting the circle first at A and C (the near points) and then at B and D (the far points), the products of the whole secant and its external part are equal:
割线是与圆有两个交点的直线。当从同一外点 P 引出两条割线，分别先交圆于 A、C（近点），再交圆于 B、D（远点）时，"整条割线 × 其圆外部分"的乘积相等：
PA × PB = PC × PD
y the external segment by the internal segment instead of by the whole secant. The proof again relies on similar triangles ACP and DBP, since ∠P is common and ∠PAB = ∠PDC (exterior angle of cyclic quadrilateral ABCD equals the opposite interior angle).
其中 PA、PC 是圆外部分（从 P 到近交点），而 PB、PD 是整条割线的长度。常见错误是用圆外部分乘以圆内部分，而不是乘以整条割线。证明同样依赖相似三角形 ACP 与 DBP：∠P 为公共角，且 ∠PAB = ∠PDC（圆内接四边形 ABCD 的外角等于内对角）。
= 5 × PD, so PD = 40 ÷ 5 = 8 cm, and the internal part CD = 8 − 5 = 3 cm. Always label which points are near and which are far before substituting.
例题：从 P 引出的第一条割线交圆于 A、B，PA = 4 cm，PB = 10 cm；第二条割线交圆于 C、D，PC = 5 cm。则 4 × 10 = 5 × PD，所以 PD = 40 ÷ 5 = 8 cm，圆内部分 CD = 8 − 5 = 3 cm。代入前务必先标清哪些点是近点、哪些是远点。
8. Tangent-Secant Theorem | 切割线定理
s a tangent touching at T and the other is a secant meeting the circle at A and B, the two "cuts" from P coincide at T, so PA × PB reduces to PT × PT:
切割线定理是上一定理的极限情形。当从外点 P 引出的一条直线是切线（切于 T），另一条是割线（交圆于 A、B）时，由 P 出发的两个"交点"在 T 处重合，于是 PA × PB 退化为 PT × PT：
PT² = PA × PB
= 9 cm and the chord AB = 9 − 4 = 5 cm. This theorem is the reason examination questions often give a tangent length and a short external segment, expecting you to find the full secant length in one line.
例题：从 P 引出的切线在 T 处与圆相切，PT = 6 cm；过 P 的割线交圆于 A、B，PA = 4 cm。则 36 = 4 × PB，所以 PB = 9 cm，弦 AB = 9 − 4 = 5 cm。正因如此，考试题常给出切线长和较短的圆外部分，期望你一步求出整条割线的长度。
nd tangent-secant - can be written as PA × PB = PC × PD once you allow the tangent to be a "double point" of a point theorem, and recognising it saves a great deal of time in multi-step problems.
请注意这一族结果的内在统一性。相交弦、两割线、切割线三种情形，只要允许切点被视为"重合的两个点"，都可以写成 PA × PB = PC × PD。这就是"点对圆的幂"定理，识别出这一点可以大幅节省多步题的时间。
9. Worked Multi-Step Problems | 多步综合例题
Examination questions rarely test one theorem alone. The following example combines tangent, radius and alternate segment ideas in the standard Edexcel IGCSE style.
考试题很少只考查单一定理。下面这道题按照 Edexcel IGCSE 的常见风格，把切线、半径与交错弧定理结合在一起。
le with centre O, and A and B are the points of contact. The chord AB is drawn. Given that ∠APB = 44°, find ∠AOB, ∠PAB and ∠ACB where C is a point on the major arc AB. Solution: since PA and PB are tangents, ∠OAP = ∠OBP = 90°, so in quadrilateral OAPB, ∠AOB = 360° − 90° − 90° − 44° = 136°. Because PA = PB, triangle APB is isosceles and ∠PAB = ∠PBA = (180° − 44°) ÷ 2 = 68°. Finally, the angle in the alternate segment gives ∠ACB = ∠PAB = 68°.
题目：PA 与 PB 是圆（圆心 O）的两条切线，A、B 为切点，连弦 AB。已知 ∠APB = 44°，求 ∠AOB、∠PAB，以及优弧 AB 上一点 C 处的 ∠ACB。解答：因 PA、PB 为切线，∠OAP = ∠OBP = 90°，故在四边形 OAPB 中，∠AOB = 360° − 90° − 90° − 44° = 136°。又因 PA = PB，三角形 APB 为等腰，∠PAB = ∠PBA = (180° − 44°) ÷ 2 = 68°。最后由交错弧定理，∠ACB = ∠PAB = 68°。
and PA = 8 cm, find AB. Using PT² = PA × PB: 144 = 8 × PB, so PB = 18 cm and AB = 18 − 8 = 10 cm. If the question then asks for the radius, and the centre O is at a perpendicular distance of 5 cm from the chord AB, use r² = 5² + (10 ÷ 2)² = 25 + 25 = 50, so r = √50 ≈ 7.07 cm.
第二类题目给出数值长度。从外点 P 引出的切线切于 T，割线交圆于 A、B。已知 PT = 12 cm，PA = 8 cm，求 AB。由 PT² = PA × PB 得 144 = 8 × PB，所以 PB = 18 cm，AB = 18 − 8 = 10 cm。若题目进一步求半径，且圆心 O 到弦 AB 的垂距为 5 cm，则用 r² = 5² + (10 ÷ 2)² = 25 + 25 = 50，得 r = √50 ≈ 7.07 cm。
10. Common Mistakes and Exam Technique | 常见错误与应试技巧
Marks in circle theorem questions are almost always split between the numerical answer and the reason given. Even with a correct value, a missing or vague reason costs a mark, so build the habit of quoting the theorem in precise wording after every step.
圆定理题目的分数几乎总是分为数值答案分与理由分两部分。即使数值正确，理由缺失或表述含糊也会扣分，因此要养成每一步之后都用准确措辞写出定理名称的习惯。
Frequent errors to avoid: mixing up the external segment with the whole secant in the secant-secant and tangent-secant theorems; assuming a line is a tangent without being told; treating the angle between the tangent and a chord as equal to the angle on the same side of the chord rather than the alternate segment; and forgetting that the two tangents from an external point are equal, which unlocks isosceles triangles and angle sums.
需要避免的常见错误：在割线-割线与切割线定理中把圆外部分与整条割线混淆；在未说明的情况下擅自认定某直线是切线；把切弦角错配为弦同侧的角而不是交错弧中的角；以及忘记从外点引出的两条切线相等，从而无法使用等腰三角形与角度和。
Standard reason statements worth memorising: "the angle between a tangent and a radius is 90°"; "tangents from an external point are equal in length"; "the alternate segment theorem"; "the perpendicular from the centre to a chord bisects the chord"; "the angle in a semicircle is 90°"; "angles in the same segment are equal"; "opposite angles of a cyclic quadrilateral sum to 180°"; and "the exterior angle of a cyclic quadrilateral equals the opposite interior angle".
值得背诵的标准理由表述："切线与半径的夹角为 90°"；"从外点引出的两条切线长度相等"；"切弦角定理（交错弧定理）"；"从圆心向弦作的垂线平分该弦"；"半圆上的圆周角为 90°"；"同弧所对的圆周角相等"；"圆内接四边形对角互补"；"圆内接四边形的外角等于内对角"。
11. Summary Table and Quick Reference | 总结表格与速查表
The table below gathers every theorem from both parts of this series into one revision sheet. Use it as a final check before the examination.
下表把本系列两篇文章中的全部定理汇总成一张复习清单，可在考前作最后检查之用。
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