Alternate Segment Theorem and Intersecting Chords: IGCSE Maths
The alternate segment theorem and the intersecting chords theorem are the final two rules in the IGCSE circle geometry set. This page walks through both with clear diagrams, worked examples, and free auto-marked practice. Use them alongside the six core circle theorems to tackle any higher-tier exam question on circle geometry.
Two more circle rules to finish the set
Once you are comfortable with the six core circle theorems, two more results complete the picture for IGCSE higher tier. They are the alternate segment theorem, which links a tangent and chord to the opposite segment, and the intersecting chords theorem, which turns two chords (or a chord and a secant) into a simple length equation.
Both rules show up in the harder Book 2 questions, and both reward careful labelling. If your diagram is neat, the theorem usually does the rest.
The alternate segment theorem
The full statement of the theorem is:
The angle between a tangent to a circle and a chord drawn from the point of contact is equal to the angle the chord makes in the alternate segment.
To spot it in a question, look for these three ingredients together:
- A tangent touching the circle at a point.
- A chord drawn from that same point.
- An inscribed angle sitting in the segment on the other side of that chord.
If all three are present, the tangent-chord angle and the inscribed angle are equal.
Worked examples: alternate segment
💡 Example 1
PT is a tangent to the circle at T. TA is a chord. Angle PTA = 38°. Find the angle TBA, where B is a point on the major arc.
Tangent PT and chord TA meet at T.
By the alternate segment theorem:
\(\angle TBA = \angle PTA\)
\(\angle TBA = 38^{\circ}\)
What's happening?
The angle between the tangent and the chord at the point of contact always equals the inscribed angle in the opposite segment. Just copy the 38° across to B.
💡 Example 2
Tangent LM touches the circle at M. Chord MN is drawn. The angle MQN in the alternate segment is 54°. Find the angle LMN.
Tangent LM and chord MN meet at M.
Angle MQN is in the alternate segment.
\(\angle LMN = \angle MQN\)
\(\angle LMN = 54^{\circ}\)
What's happening?
The direction of the rule works both ways. Whichever angle you know (at the tangent or in the alternate segment), the other one matches.
Intersecting chords theorem
When two chords cross each other inside a circle, the products of the pieces are equal.
If chords AB and CD meet at P inside the circle, then \(AP \times PB = CP \times PD\).
This turns a geometric picture into a straightforward length equation. Label the four short segments, substitute, solve.
When the chords meet outside the circle
If two lines from an external point P each cut the circle at two places (two secants), a similar rule applies:
\(PA \times PB = PC \times PD\)
Here PA and PB are measured along the same secant from P to the two points where it crosses the circle. Same for PC and PD along the second secant. If one line is a tangent touching at a single point T, and the secant cuts the circle at A and B, the rule becomes \(PT^{2} = PA \times PB\).
Worked examples: intersecting chords
💡 Example 3
Chords AB and CD cross at P inside a circle. AP = 6 cm, PB = 4 cm, CP = 3 cm. Find PD.
Apply the intersecting chords rule:
\(AP \times PB = CP \times PD\)
\(6 \times 4 = 3 \times PD\)
\(24 = 3 \times PD\)
\(PD = 8\) cm
What's happening?
Substitute the three known lengths, multiply out, then divide. Always the same four-step pattern: identify the four pieces, write the equation, substitute, solve.
💡 Example 4
Two secants drawn from an external point P cross a circle. The first passes through points A and B with PA = 5 cm and AB = 7 cm. The second passes through C and D with PC = 4 cm. Find PD.
PB is the whole distance from P to B:
\(PB = PA + AB\)
\(PB = 5 + 7\)
\(PB = 12\) cm
Apply the two-secants rule:
\(PA \times PB = PC \times PD\)
\(5 \times 12 = 4 \times PD\)
\(60 = 4 \times PD\)
\(PD = 15\) cm
What's happening?
The trap is using AB directly. The rule needs the full distance from P to the far point, so add PA and AB first, then substitute.
💡 Example 5: tangent and secant from the same point
From an external point P, a tangent PT touches a circle at T, and a secant from P cuts the circle at A (nearer to P) and B (farther from P). If PT = 8 cm and PA = 4 cm, find the length PB.
Apply the tangent-secant rule:
\(PT^{2} = PA \times PB\)
\(8^{2} = 4 \times PB\)
\(64 = 4 \times PB\)
\(PB = 16\) cm
The chord AB has length:
\(AB = PB - PA\)
\(AB = 16 - 4\)
\(AB = 12\) cm
What's happening?
The tangent length gets squared on one side; the product of the near and far distances along the secant gives the other. Once PB is known, the chord itself drops out by subtracting PA.
🔑 Key points
- Alternate segment: the angle between the tangent and a chord at the point of contact equals the inscribed angle in the opposite segment.
- Two chords inside: if AB and CD meet at P inside the circle, \(AP \times PB = CP \times PD\).
- Two secants outside: from external point P, \(PA \times PB = PC \times PD\), where each product uses the whole distance from P to the far point.
- Tangent and secant: \(PT^{2} = PA \times PB\) when a tangent PT and a secant cutting the circle at A and B come from the same external point.
- Always state the theorem by name when giving reasons in exam answers.
⚠️ Common pitfalls
- Picking the wrong segment. The alternate segment is on the opposite side of the chord from the tangent-chord angle.
- Using the internal chord length AB instead of the full distance PB from the external point to the far intersection.
- Forgetting that the theorem only works when the line touches the circle. A tangent meets the circle at exactly one point.
- Squaring the wrong length for the tangent-secant case. The tangent length is the one that gets squared, not the secant length.
- Giving a numerical answer without naming the theorem. Edexcel mark schemes reward the reason as often as the number.