MathsICSE Class 10 · Tangent and Secant Properties

Intersecting Chords and the Tangent–Secant Theorem: PA × PB = PC × PD and PT² = PA × PB (ICSE Class 10)

Q: What is the intersecting chords theorem?

If two chords of a circle, or the lines containing them, meet at a point P, the product of the two segments of one chord, measured from P, equals the product for the other: PA × PB = PC × PD.

Q: What is the tangent–secant theorem?

If PT is a tangent and PAB is a secant from the same external point P, then PT² = PA × PB, where PB is the whole distance from P to the far point.

Q: Do I need to prove these theorems for ICSE?

The syllabus says proofs of these theorems are to be taught, and 2025 Q10(ii) asked for the similar-triangles step before the length. Learn the AA similarity proofs.

Updated 15 October 2026 · Fact-checked

If two chords $AB$ and $CD$ meet at $P$, inside or outside the circle, then $PA \times PB = PC \times PD$. If $PT$ is a tangent and $PAB$ a secant from an external point $P$, then $PT^2 = PA \times PB$. Every length is measured from $P$, and both results follow from similar triangles.

Every length starts at P: PA × PB = PC × PD for chords, and the tangent is the limiting case so PT² = PA × PB.

Illustration drawn for QuestionBuddy.

Understand Intersecting Chords and the Tangent–Secant Theorem

The CISCE syllabus lists two product theorems under Tangent and Secant Properties; their proofs are to be taught.

1. Intersecting chords (internally or externally). If chords $A B$ and $C D$ (or the lines containing them) meet at $P$ , then $P A \times P B = P C \times P D$ .

*Proof, $P$ inside.* Join $A C$ and $B D$ . In $△ P A C$ and $△ P D B$ : $∠ A P C = ∠ D P B$ (vertically opposite) and $∠ C A B = ∠ C D B$ (angles in the same segment, chord $C B$ ). So $△ P A C \sim △ P D B$ (AA), giving $\frac{P A}{P D} = \frac{P C}{P B}$ , that is $P A \times P B = P C \times P D$ .

*Proof, $P$ outside* (secants $P A B$ and $P C D$ , $A$ and $C$ nearer to $P$ ). $∠ P$ is common, and $∠ P A C = ∠ P D B$ (the exterior angle $∠ P A C$ of cyclic quadrilateral $A B D C$ equals the interior opposite angle $∠ B D C$ ). So $△ P A C \sim △ P D B$ and the same product follows.

2. Tangent and secant. If $P T$ touches the circle at $T$ and $P A B$ is a secant, then $P T^{2} = P A \times P B$ .

*Proof.* In $△ P T A$ and $△ P B T$ : $∠ P$ is common and $∠ P T A = ∠ P B T$ (alternate segment theorem). So $△ P T A \sim △ P B T$ and $\frac{P T}{P B} = \frac{P A}{P T}$ , giving $P T^{2} = P A \times P B$ . The tangent is the limiting case of a secant whose two points have come together at $T$ .

**The golden rule: measure from $P$ .** $P A$ and $P B$ are both distances from the meeting point $P$ , not the chord length $A B$ . For an external point, $P B = P A + A B$ .

Useful special cases.

Secant through the centre: if $P A$ is the near part and $r$ the radius, $P B = P A + 2 r$ ; you can check with $P T^{2} = O P^{2} - r^{2}$ .
Arch or broken plate: a chord of length $2 c$ with height (rise) $h$ above its mid-point gives $c \times c = h \times (2 r - h)$ , so $2 r = \frac{c ^{2}}{h} + h$ .
Equal tangents from a common secant: a point on line $A B$ produced, where $A B$ is the common chord of two circles, has equal tangents to both circles.

In the papers. 2024 Q9(i) (3 marks) used $P Q^{2} = P A \times P B$ with a diameter produced; 2025 Q10(ii) (3 marks) asked candidates to prove the similarity first and then find $P T$ .

Intersecting Chords and the Tangent–Secant Theorem: formulas to remember

Intersecting chords: $P A \times P B = P C \times P D$; $P$ inside or outside the circle; lengths from $P$ .
Tangent–secant: $P T^{2} = P A \times P B$; $P T$ tangent, $P A B$ secant, $P B = P A + A B$ .
Rise and span: $2 r = \frac{c ^{2}}{h} + h$; Chord $2 c$ , rise $h$ at its mid-point.

How to solve Intersecting Chords and the Tangent–Secant Theorem questions

Find the meeting point $P$ , write every length from $P$ , then multiply.

1Locate $P$ , where the two chords, secants or the tangent and secant meet.
2Label each segment from $P$ ; for an external point, add the chord to the near part to get the far part.
3Write $P A \times P B = P C \times P D$ or $P T^{2} = P A \times P B$ .
4Solve; if an unknown appears in both factors, you get a quadratic: reject negative roots.
5If the question says 'prove' or 'show', first prove the similar triangles with two pairs of equal angles and reasons, then write the ratio of corresponding sides.
6Check with Pythagoras when the secant passes through the centre.

Intersecting Chords and the Tangent–Secant Theorem in real life

Each concept, met in an everyday situation. Read the situation, try the problem, then check the solution. The one-line hook is what to remember.

Intersecting chords in a broken plate

Suppose a piece of a broken circular plate has a straight edge (a chord) 30 cm long, and the curved edge rises 9 cm above the mid-point of that chord.

Find the diameter of the original plate.

Try it first, then show the solutionSolution

Extend the 9 cm height through the centre to make a diameter. It cuts the chord into 15 cm and 15 cm.
Intersecting chords: $15 \times 15 = 9 \times (d - 9)$ , so $d - 9 = 25$ .

Answer: 34 cm

Memory hook: Half-chord squared = rise × the rest of the diameter.

Tangent and secant from one point

Imagine a gate $P$ near a circular pond. One straight path from $P$ just touches the pond at $T$ , and $P T = 20$ m. Another straight path from $P$ meets the pond's edge after 16 m and goes straight across it.

How far does the second path run across the pond?

Try it first, then show the solutionSolution

$P T^{2} = P A \times P B$ : $400 = 16 \times P B$ , so $P B = 25$ m.
Distance across $= P B - P A = 25 - 16$ .

Answer: 9 m

Memory hook: Tangent squared = near × far.

Tricks to remember Intersecting Chords and the Tangent–Secant Theorem

Short memory aids. Say each one aloud once, then use it on a question the same day.

Near × far
From the point $P$ : near part × far part is the same on every line through $P$ ; for a tangent, near = far = $P T$ .
Use it when: Writing the right product every time.
Measure from P
If a length does not start at $P$ , it does not go in the formula.
Use it when: Avoiding $P A \times A B$ errors.
Proof pattern
Common or vertically opposite angle + one circle-theorem angle = AA similarity.
Use it when: Every 'prove that' version of these theorems.

Intersecting Chords and the Tangent–Secant Theorem: solved examples

Ordered from board level to HOTS. Try each one before you open the solution.

Example 1: Chords crossing inside · 1 mark · MCQBoard level

Chords $A B$ and $C D$ of a circle meet at $P$ inside it. $P A = 3$ cm, $P B = 8$ cm and $P C = 4$ cm. Then $P D$ is:
(a) 6 cm
(b) 9 cm
(c) 10.7 cm
(d) 12 cm

Try it first, then show the solutionSolution

$P A \times P B = P C \times P D$ : $24 = 4 \times P D$ , so $P D = 6$ cm.

Answer: (a)

Example 2: Tangent from the secant · 2 marks · Short answerBoard level

From an external point $P$ , a tangent $P T$ and a secant $P A B$ are drawn to a circle. $P A = 3$ cm and $A B = 24$ cm. Find $P T$ .

Try it first, then show the solutionSolution

$P B = P A + A B = 27$ cm.
$P T^{2} = P A \times P B = 3 \times 27 = 81$ , so $P T = 9$ cm.

Answer: 9 cm

Example 3: Two secants from one point · 3 marks · Short answerHarder board level

$P A B$ and $P C D$ are two secants of a circle from an external point $P$ . $P A = 5$ cm, $A B = 7$ cm and $P C = 4$ cm. Find $P D$ and $C D$ .

Try it first, then show the solutionSolution

$P B = 5 + 7 = 12$ cm.
$P A \times P B = P C \times P D$ (chords intersecting externally): $5 \times 12 = 4 \times P D$ , so $P D = 15$ cm.
$C D = P D - P C = 15 - 4 = 11$ cm.

Answer: $P D = 15$ cm, $C D = 11$ cm

Example 4: Proof and a quadratic · 4 marks · Long answerHarder board level

(a) Chords $A B$ and $C D$ of a circle intersect at $P$ inside it. Prove that $P A \times P B = P C \times P D$ .
(b) $A B = 14$ cm with $P A = 6$ cm, and $C D = 16$ cm. Find $P C$ and $P D$ , given $P C < P D$ .

Try it first, then show the solutionSolution

(a) Join $A C$ and $B D$ . In $△ P A C$ and $△ P D B$ : $∠ A P C = ∠ D P B$ (vertically opposite); $∠ C A P = ∠ B D P$ (angles in the same segment on chord $C B$ ).
So $△ P A C \sim △ P D B$ (AA), and $\frac{P A}{P D} = \frac{P C}{P B}$ , i.e. $P A \times P B = P C \times P D$ .
(b) $P B = 8$ cm, so $P C \times P D = 48$ with $P C + P D = 16$ .
Let $P C = x$ : $x (16 - x) = 48$ , so $x^{2} - 16 x + 48 = 0$ , $(x - 4) (x - 12) = 0$ .
$P C = 4$ cm and $P D = 12$ cm.

Answer: (b) $P C = 4$ cm, $P D = 12$ cm

Example 5: Proving the tangent–secant theorem · 4 marks · Long answerHOTS / competency

A tangent at $T$ meets the chord $S R$ produced at $P$ , with $R$ between $P$ and $S$ .
(a) Prove that $△ P T R \sim △ P S T$ .
(b) Hence show that $P T^{2} = P R \times P S$ .
(c) If $P R = 5$ cm and $R S = 40$ cm, find $P T$ and the ratio of the areas of $△ P T R$ and $△ P S T$ .

Try it first, then show the solutionSolution

(a) $∠ T P R = ∠ S P T$ (common). $∠ P T R = ∠ P S T$ (angle between tangent $P T$ and chord $T R$ equals the angle in the alternate segment). So $△ P T R \sim △ P S T$ (AA).
(b) Corresponding sides: $\frac{P T}{P S} = \frac{P R}{P T}$ , so $P T^{2} = P R \times P S$ .
(c) $P S = 45$ cm, $P T^{2} = 5 \times 45 = 225$ , $P T = 15$ cm.
Area ratio $= (\frac{P T}{P S})^{2} = (\frac{15}{45})^{2} = \frac{1}{9}$ , i.e. $1 : 9$ .

Answer: (c) $P T = 15$ cm; areas $1 : 9$

Example 6: Case study: a circular arch bridge · 4 marks · Case-basedHOTS / competency

The underside of a bridge is an arc of a circle. The span (the chord at water level) is 24 m and the highest point of the arch is 6 m above the water.
(a) Find the radius of the arc.
(b) How far below the water level is the centre?
(c) Find the height of the arch above a point on the water 6 m from one end of the span, to 2 decimal places.

Try it first, then show the solutionSolution

(a) The vertical through the top is a diameter; it cuts the span into 12 m and 12 m. $12 \times 12 = 6 \times (2 r - 6)$ , so $2 r - 6 = 24$ and $r = 15$ m.
(b) Centre below water $= r - 6 = 9$ m.
(c) The point is 6 m from the mid-point. The vertical chord through it is $6$ m from the centre's vertical, so its upper end is $1 5^{2} - 6^{2} = 189 \approx 13.75$ m above the centre.
Height above water $= 189 - 9 \approx 4.75$ m. Check by chords: the point divides the span into 6 m and 18 m, and $4.75 \times (4.75 + 18) \approx 108 = 6 \times 18$ ✓

Answer: (a) 15 m (b) 9 m (c) about 4.75 m

Where students lose marks in Intersecting Chords and the Tangent–Secant Theorem

The slips examiners see most often, and what to write instead.

Mistake: Using the chord instead of the distance from P
Both factors are measured from $P$ ; the far part is $P A + A B$ .
Wrong working
$P T^{2} = P A \times A B$ .
Right working
$P T^{2} = P A \times (P A + A B)$ .
Do this: Write $P B = P A + A B$ on its own line first.
Usually costs: Reported in 2024 Q9(i): candidates used $PQ^2 = PO \times PB$ or $AB \times BP$.
Mistake: Radius instead of diameter
When a diameter is produced, the near part is the far part minus the whole diameter.
Wrong working
$P A = 24$ , $r = 9$ : $P B = 24 - 9 = 15$ .
Right working
$P B = 24 - 18 = 6$ , $P Q^{2} = 24 \times 6 = 144$ , $P Q = 12$ cm.
Do this: Mark the diameter's length on the figure.
Usually costs: Reported in 2024 Q9(i): candidates subtracted the radius 9 instead of the diameter 18.
Mistake: Quoting the result in a 'prove' question
When the question asks for the similarity, the product must be derived from it.
Wrong working
' $P T^{2} = P R \times P S$ (tangent–secant theorem)' with no similar triangles.
Right working
Prove $△ P T R \sim △ P S T$ with $∠ P$ common and $∠ P T R = ∠ P S T$ , then cross-multiply.
Do this: Name the equal angles with three letters and give reasons.
Usually costs: Reported in 2025 Q10(ii): candidates wrote $\angle T = \angle S$ or stated the result directly.

Exam tips

2024 Q9(i) (3 marks): radius 9 cm, $P A = 24$ cm: $P B = 6$ cm, $P Q = 12$ cm; Pythagoras with $O P = 15$ cm is an acceptable second method.
2025 Q10(ii) (3 marks): prove $△ P T R \sim △ P S T$ (alternate segment theorem), then $P T^{2} = P R \times P S = 4 \times 16$ , $P T = 8$ cm.
Write corresponding vertices in the same order when stating similarity; the ratio of sides then reads off directly.
Reject negative roots when the unknown length appears in both factors.

Practice questions on Intersecting Chords and the Tangent–Secant Theorem

Write each answer on paper first. Open the hint only if you are stuck, and the answer only when you have finished.

Q1 · 1 mark · MCQBoard level
$P T$ is a tangent and $P A B$ a secant of a circle. $P T = 15$ cm and $P A = 9$ cm. Then $P B$ is:
(a) 6 cm
(b) 16 cm
(c) 24 cm
(d) 25 cm
Hint
$P T^{2} = P A \times P B$ .
Answer
(d)
Q2 · 2 marks · Short answerBoard level
Chords $A B$ and $C D$ meet at $P$ inside a circle. $A P = 2.5$ cm, $P B = 4$ cm and $C P = 2$ cm. Find $P D$ and $C D$ .
Hint
$A P \times P B = C P \times P D$ .
Answer
$P D = \frac{10}{2} = 5$ cm; $C D = 7$ cm.
Q3 · 1 mark · Assertion-reasonHarder board level
Assertion (A): If secants $P A B$ and $P C D$ meet at an external point $P$ , then $P A \times A B = P C \times C D$ .
Reason (R): $△ P A C \sim △ P D B$ .
(a) (A) is true and (R) is false.
(b) (A) is false and (R) is true.
(c) Both (A) and (R) are true, and (R) is the correct explanation for (A).
(d) Both (A) and (R) are true, but (R) is not the correct explanation for (A).
Hint
Which lengths does the similarity actually give?
Answer
(b): the correct result is $P A \times P B = P C \times P D$ .
Q4 · 3 marks · Short answerHarder board level
A secant from an external point $P$ passes through the centre $O$ of a circle of radius 8 cm and meets the circle first at $A$ , with $P A = 2$ cm. Find $P B$ , where $B$ is the other end, and the length of the tangent from $P$ . Check your answer using $O P$ .
Hint
$P B = P A + 2 r$ .
Answer
$P B = 18$ cm; $P T = 2 \times 18 = 6$ cm. Check: $O P = 10$ , $1 0^{2} - 8^{2} = 6$ ✓
Q5 · 3 marks · Short answerHarder board level
$P A B$ and $P C D$ are secants from an external point $P$ . $P A = 6$ cm, $P B = 15$ cm and $C D = 13$ cm. Find $P C$ and $P D$ .
Hint
Let $P C = x$ ; then $P D = x + 13$ .
Answer
$x (x + 13) = 90$ , $(x + 18) (x - 5) = 0$ , so $P C = 5$ cm and $P D = 18$ cm.
Q6 · 3 marks · Short answerHarder board level
From an external point $P$ , a tangent $P T = 6$ cm and a secant $P A B$ are drawn. $A B = 5$ cm. Find $P A$ .
Hint
$P B = P A + 5$ .
Answer
$x (x + 5) = 36$ , $(x + 9) (x - 4) = 0$ , so $P A = 4$ cm.
Q7 · 4 marks · Long answerHOTS / competency
Two circles intersect at $A$ and $B$ . $P$ is a point on $B A$ produced, and $P T$ and $P S$ are tangents from $P$ to the two circles.
(a) Prove that $P T = P S$ .
(b) If $P A = 4$ cm and $A B = 5$ cm, find $P T$ .
Hint
$P A B$ is a secant of both circles.
Answer
(a) For the first circle $P T^{2} = P A \times P B$ ; for the second $P S^{2} = P A \times P B$ . So $P T^{2} = P S^{2}$ and $P T = P S$ . (b) $P T = 4 \times 9 = 6$ cm.

Test your recall

Answer in your head or on paper first, then open the card. Pulling an answer out of memory makes it stick far better than re-reading.

1.Chords $A B$ , $C D$ meet at $P$ . Relation?Show answer
$P A \times P B = P C \times P D$
2.Tangent $P T$ , secant $P A B$ . Relation?Show answer
$P T^{2} = P A \times P B$
3. $P A = 2$ , $P B = 6$ , $P C = 3$ . $P D$ ?Show answer
4
4.Which theorem gives the equal angle in the tangent–secant proof?Show answer
The alternate segment theorem.
5.Which triangles are similar when chords cross inside at $P$ ?Show answer
$△ P A C$ and $△ P D B$ .

Intersecting Chords and the Tangent–Secant Theorem: frequently asked questions

What is the intersecting chords theorem?

If two chords of a circle, or the lines containing them, meet at a point P, the product of the two segments of one chord, measured from P, equals the product for the other: PA × PB = PC × PD.

What is the tangent–secant theorem?

If PT is a tangent and PAB is a secant from the same external point P, then PT² = PA × PB, where PB is the whole distance from P to the far point.

Do I need to prove these theorems for ICSE?

The syllabus says proofs of these theorems are to be taught, and 2025 Q10(ii) asked for the similar-triangles step before the length. Learn the AA similarity proofs.