Geometry

Segments of chord and formed rectangle

The intersecting chords theorem and its associated rectangle properties form a cornerstone of Euclidean circle geometry, establishing a powerful algebraic product relationship between chord segments. This topic covers this concept.

For: UP PGT Mathematics UP TGT Mathematics DSSSB PGT / TGT KVS / NVS PGT LT Grade

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Definition and statement

Intersecting Chords Theorem (Theorem of Chords / Rectangle Formed by Chord Segments)

If two chords $AB$ and $CD$ of a circle intersect at a point $P$ (either internally inside the circle or externally when extended outside), then the rectangle contained by the segments of one chord is equal in area to the rectangle contained by the segments of the other chord. That is, $$PA \cdot PB = PC \cdot PD$$

📐 Geometric Interpretation

Geometrically, when two chords intersect, they divide each other into two directed line segments. Constructing rectangles with adjacent sides equal to the respective segments of each chord yields two rectangles of identical area. This is deeply tied to the concept of power of a point with respect to a circle and cyclic quadrilaterals formed by joining the endpoints.

⚙️ Algebraic & Structural Meaning

Algebraically, it establishes a quadratic invariant relationship among the segment lengths. If $P$ divides chord $AB$ into segments of length $x$ and $y$, and chord $CD$ into segments of length $u$ and $v$, then $xy = uv$. This proportionality allows the determination of unknown segment lengths using simple algebraic equations.

Conditions and hypotheses

Check each condition and what changes when it is omitted:

Hypothesis / Condition	Domain / Interval	Necessity	What Fails If Condition Dropped
Concyclicity of Endpoints	`Circular Arc / Concyclic Points`	Strictly Mandatory / Necessary	If the four points do not lie on the same circle, the property of intersecting chords fails, and $PA \cdot PB \neq PC \cdot PD$ in general (it only holds when the points are concyclic).
Intersection Point $P$	`Interior or Exterior Point in the plane of the circle`	Strictly Mandatory	If lines containing chords $AB$ and $CD$ are parallel or do not intersect, the product relation for intersecting chords is undefined.
Collinearity and Segment Orientation	`Directed segments along secant lines`	Essential for correct sign and magnitude	Ignoring the correct relative position of point $P$ relative to the chord endpoints leads to sign errors in external secant configurations.

Useful results and shortcuts

Use these identities and results when their conditions apply:

Internal Intersection Product Rule

$$AP \cdot PB = CP \cdot PD$$

Rule: If point $P$ lies inside the circle on chords $AB$ and $CD$, simply equate the product of the two parts of the first chord to the product of the two parts of the second chord.

Application: Directly tested in UP PGT geometry numerical problems to find an unknown segment length.

External Secant-Chord (Tangent-Secant) Rule

$$PT^2 = PA \cdot PB$$

Rule: If a tangent segment $PT$ and a secant $PAB$ are drawn from an external point $P$, the square of the tangent length equals the product of the external secant segment and the entire secant length.

Perpendicular Chord from Center

$$AM = MB = \frac{1}{2}AB = \sqrt{R^2 - d^2}$$

Rule: Perpendicular from the center bisects the chord. Use right triangle formed by radius $R$, distance from center $d$, and half-chord $AM$ to setup quick algebraic rectangles.

Power of a Point Formula

$$PA \cdot PB = |d^2 - R^2$$

Rule: The power of a point $P$ at distance $d$ from the center of a circle of radius $R$ is constant for any chord passing through $P$, given by $d^2 - R^2$.

Application: Used to eliminate lengthy chord calculations in advanced geometry questions.

Equal Chords Equidistant from Center

$$AB = CD \iff d_1 = d_2$$

Rule: Chords of equal length are equidistant from the center; hence their segment rectangles are symmetric.

Application: Enables rapid option elimination in structural geometry questions.

Rapid Ratio & Area Elimination

$$ \frac{Area(\triangle APC)}{Area(\triangle DPB)} = \frac{AP \cdot CP}{DP \cdot BP} $$

Rule: Use similar triangles $\triangle APC \sim \triangle DPB$ (by subtended angles in same segment) to establish rapid side ratios without full vector/coordinate geometry.

Application: Saves 60+ seconds in multiple-choice assertion-reason questions.

Common errors and counterexamples

These examples show how a conclusion can fail when a condition is missing:

⚠️ COMMON ERROR #01

Confusing Internal vs. External Intersection Segments

Common Pitfall: Candidates use the entire chord length instead of the specific segment from the intersection point $P$ to the circle boundary.

Counterexample: For chord $AB$ divided into $AP$ and $PB$, using $AB$ instead of $AP$ or $PB$.

Why it fails: The theorem strictly states that the product of the parts into which the chord is divided by the intersection point are equal, not the whole chords.

🛡️ Exam Defense: Always measure segments starting from the intersection point $P$ to the endpoints $A, B, C, D$.

⚠️ COMMON ERROR #02

Misapplying External Secant Lengths

Common Pitfall: In external intersection ($P$ outside), taking the product of external part and internal part instead of external part and total secant length.

Counterexample: $PA \cdot AB$ instead of $PA \cdot PB$ where $PB$ is the total length from $P$ to the far end $B$.

Why it fails: The secant theorem involves the external segment and the *entire* secant line segment.

🛡️ Exam Defense: For external points, remember: (External segment) $\times$ (Total secant) = constant.

Comparison

Concept / Theorem	Hypotheses / Assumptions	Key Formula / Outcome	Geometric Meaning	Exam Identification Clue
Internal Intersecting Chords	Point $P$ inside circle, chords $AB$ and $CD$ intersect at $P$	$PA \cdot PB = PC \cdot PD$	Rectangles formed by chord segments are equal in area	Keywords: ' chords intersect internally at point P '
External Intersecting Chords (Secants)	Point $P$ outside circle, secants $PAB$ and $PCD$	$PA \cdot PB = PC \cdot PD$	Power of point from exterior equals secant products	Keywords: ' produced chords meet outside at P '
Tangent-Secant Theorem	Point $P$ outside, tangent $PT$ and secant $PAB$	$PT^2 = PA \cdot PB$	Square of tangent equals external secant rectangle	Keywords: ' tangent and secant drawn from external point '
Perpendicular Bisector of Chord	Center $O$ and chord $AB$	$AM = MB, OM \perp AB$	Radius perpendicular to chord bisects it	Keywords: ' perpendicular from center to chord '

Solved examples

Each problem includes a full solution and a shorter method:

CANONICAL PROBLEM 01 Dual-Method Breakdown

Two chords $AB$ and $CD$ of a circle intersect internally at a point $P$. If $PA = 4\text{ cm}$, $PB = 6\text{ cm}$, and $PC = 3\text{ cm}$, find the length of $PD$.

(A) $5\text{ cm}$(B) $8\text{ cm}$(C) $6\text{ cm}$(D) $10\text{ cm}$

📚 Method A: Formal Textbook Derivation

Step 1: Apply the internal intersecting chords theorem: $PA \cdot PB = PC \cdot PD$. Step 2: Substitute given values: $4 \times 6 = 3 \times PD$. Step 3: Solve for $PD$: $24 = 3 \cdot PD \implies PD = 8\text{ cm}$.

⚡ Method B: 25-Second Exam Shortcut

Mental math: $4 \times 6 = 24$, divide by $3$ gives $8$. Done in 5 seconds.

💡 Tip: Verify that $P$ is indeed internal by checking that segment lengths sum to less than diameter if needed, but direct formula applies.

CANONICAL PROBLEM 02 Dual-Method Breakdown

From an external point $P$, a secant $PAB$ is drawn such that $PA = 4\text{ cm}$ and $AB = 5\text{ cm}$. If a tangent $PT$ is drawn from $P$ to the same circle, find the length of $PT$.

(A) $4\text{ cm}$(B) $5\text{ cm}$(C) $6\text{ cm}$(D) $3\text{ cm}$

📚 Method A: Formal Textbook Derivation

Step 1: Identify total secant length $PB = PA + AB = 4 + 5 = 9\text{ cm}$. Step 2: Apply tangent-secant theorem: $PT^2 = PA \cdot PB$. Step 3: Substitute values: $PT^2 = 4 \times 9 = 36$. Step 4: Take square root: $PT = 6\text{ cm}$.

⚡ Method B: 25-Second Exam Shortcut

Note total secant $PB = 4+5 = 9$. Then $PT = \sqrt{4 \times 9} = \sqrt{36} = 6$.

💡 Tip: Examiners frequently trap students by giving $AB = 5$ instead of total $PB = 5$. Always compute total secant length first!

Practice questions

Choose an answer to check it and see the solution:

MCQ 1: If two chords $AB$ and $CD$ intersect at right angles inside a circle, and the segments of $AB$ are $3$ and $4$, while one segment of $CD$ is $2$, what is the length of the other segment of $CD$?

A $5$

B $6$

C $4$

D $3$

Derivation & Analysis:

By intersecting chords theorem, $3 \times 4 = 2 \times x \implies 12 = 2x \implies x = 6$.

MCQ 2: A chord $AB$ of length $16\text{ cm}$ is at a distance of $6\text{ cm}$ from the center of the circle. What is the radius of the circle?

A $8\text{ cm}$

B $10\text{ cm}$

C $12\text{ cm}$

D $14\text{ cm}$

Derivation & Analysis:

Perpendicular from center bisects chord, so half-chord $AM = \frac{16}{2} = 8$. Radius $R = \sqrt{AM^2 + d^2} = \sqrt{8^2 + 6^2} = \sqrt{64 + 36} = \sqrt{100} = 10\text{ cm}$.

MCQ 3: If a point $P$ is at distance $d$ from the center of a circle of radius $R$, the power of the point $P$ with respect to the circle when $P$ is inside is:

A $R^2 - d^2$

B $d^2 - R^2$

C $R^2 + d^2$

D $\frac{R^2}{d^2}$

Derivation & Analysis:

For an interior point, the power of the point is defined as $R^2 - d^2$, which represents the product of the chord segments passing through $P$.

MCQ 4: Two intersecting chords $AB$ and $CD$ meet at $P$ inside a circle. If $AP : PB = 2 : 3$ and $CP : PD = 1 : 6$, and $AB = 10$, find $CD$.

A $14$

B $21$

C $28$

D $35$

Derivation & Analysis:

$AP = 4, PB = 6$. Product $AP \cdot PB = 24$. Thus $CP \cdot PD = 24$. Since $CP:PD = 1:6$, let $CP = x, PD = 6x$, so $6x^2 = 24 \implies x^2 = 4 \implies x = 2$. Then $CD = x + 6x = 7x = 7(2) = 14$ ... wait, $CP=2, PD=12$, so $CD = 2 + 12 = 14$. Let's check options: Option A is 14.

MCQ 5: If the radius of a circle is $13\text{ cm}$ and the length of a chord is $24\text{ cm}$, the distance of the chord from the center is:

A $5\text{ cm}$

B $7\text{ cm}$

C $12\text{ cm}$

D $6\text{ cm}$

Derivation & Analysis:

Half chord $AM = 12$. Distance $d = \sqrt{R^2 - AM^2} = \sqrt{13^2 - 12^2} = \sqrt{169 - 144} = \sqrt{25} = 5\text{ cm}$.

MCQ 6: In a circle, if two equal chords intersect, the segments of one chord are respectively equal to:

A The segments of the other chord in reverse or direct correspondence

B Half of the chord length

C Double the chord length

D Zero

Derivation & Analysis:

By symmetry, equal chords intersecting at the same angle have corresponding segments equal.

MCQ 7: A tangent $PT$ of length $12\text{ cm}$ is drawn from an external point $P$ to a circle. If a secant $PAB$ passes through the center or is at distance such that external segment $PA = 8\text{ cm}$, what is the total secant length $PB$?

A $12\text{ cm}$

B $16\text{ cm}$

C $18\text{ cm}$

D $20\text{ cm}$

Derivation & Analysis:

$PT^2 = PA \cdot PB \implies 12^2 = 8 \cdot PB \implies 144 = 8 \cdot PB \implies PB = \frac{144}{8} = 18\text{ cm}$.

MCQ 8: Assertion (A): The rectangle formed by the segments of two intersecting chords of a circle are equal in area. Reason (R): Triangles formed by joining the endpoints of the chords are similar.

A Both (A) and (R) are true and (R) is the correct explanation of (A)

B Both (A) and (R) are true but (R) is NOT the correct explanation of (A)

C (A) is true but (R) is false

D (A) is false but (R) is true

Derivation & Analysis:

Similarity of triangles $\triangle APC \sim \triangle DPB$ gives $\frac{AP}{DP} = \frac{PC}{PB} \implies AP \cdot PB = PC \cdot DP$, proving both assertion and reason.

MCQ 9: If chord $AB$ and chord $CD$ intersect at $90^\circ$ at an interior point $P$, and $AP=2, PB=8, CP=4$, then $PD$ equals:

A $2$

B $4$

C $6$

D $8$

Derivation & Analysis:

$AP \cdot PB = CP \cdot PD \implies 2 \times 8 = 4 \times PD \implies 16 = 4 \cdot PD \implies PD = 4$.

Q10

MCQ 10: Which of the following conditions is strictly necessary for the intersecting chords theorem $PA \cdot PB = PC \cdot PD$ to hold?

A The chords must be equal in length

B The four endpoints must be concyclic

C The intersection point must be the center of the circle

D The chords must be perpendicular

Derivation & Analysis:

Concyclicity of the four endpoints $A, B, C, D$ on a single circle is the fundamental geometric hypothesis required for the theorem.

Common questions

No. The relationship holds whether the intersection is internal or external. For external intersections (secants), the product of the external segment and the entire secant is constant ($PA \cdot PB = PC \cdot PD$), which is an extension of the chord rectangle principle.

Students frequently confuse the whole chord length with the segment length from the intersection point $P$, or they miscalculate external secant lengths by forgetting to add the inner chord segment to the external part.

Memorize the standard formulas ($PA \cdot PB = PC \cdot PD$ and $PT^2 = PA \cdot PB$), draw rough proportional sketches immediately on the question booklet, and utilize mental arithmetic for simple products instead of setting up full simultaneous equations.

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