Basic Proportionality Theorem

Basic Proportionality Theorem (BPT)

1.0Master Thales' Theorem in Minutes

Discover the foundational geometric theorem used to compute scaled proportions and structural heights. Learn the step-by-step mathematical derivation of the Basic Proportionality Theorem, understand its logical converse, and explore how to use its algebraic formulas to ace Class 10 board exam questions.

Class: 10 Mathematics (CBSE)
Chapter: Triangles
Estimated Learning Time: 25–30 Minutes

2.0Learning Outcomes

After completing this lesson, you will be able to:

State the Basic Proportionality Theorem (BPT) or Thales' Theorem clearly.
Set up construction elements and derive the full geometric proof using triangle areas.
State and apply the Converse of BPT to verify parallel lines in geometric figures.
Identify alternate mathematical variations of the BPT ratio formula.
Solve algebraic linear and quadratic equations based on proportional side segments.

Have you ever wondered how ancient scholars calculated the heights of massive structures, like the Great Pyramids of Giza, long before modern laser measuring tools or GPS existed? Around 600 BC, a Greek mathematician named Thales of Miletus did exactly that. He used a simple trick involving shadows, measuring the length of a stick’s shadow and comparing it to the pyramid's shadow at the exact same time of day.

The mathematical backbone of his discovery is what we study today as the Basic Proportionality Theorem (BPT), or simply Thales's Theorem.

BPT is a fundamental cornerstone of Euclidean geometry, particularly in the study of similar triangles. It acts as a gateway to advanced geometric concepts, trigonometry, and coordinate geometry. Whether you are a high school student navigating board exams or a competitive exam aspirant, mastering BPT will give you an intuitive grasp of geometric scaling and proportionality.

In this comprehensive educational guide, we will look at the core statement of BPT, break down its step-by-step geometric proof, explore its inverse (the converse), and dive into multiple solved algebraic examples.

3.0The Statement of Basic Proportionality Theorem (BPT)

The theorem establishes a strict proportional relationship between the line segments created when a parallel line cuts through a triangle.

The BPT Statement:

If a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points, then the other two sides are divided in the same ratio.

To look at this mathematically, let us consider a triangle ABC:

Suppose a straight line DE is drawn such that it is strictly parallel to the base side BC $(D E ∥ BC)$
This line intersects the side AB at point D and the side AC at point E.

According to the Basic Proportionality Theorem, this structural arrangement guarantees that:

$\frac{A D}{D B} = \frac{A E}{EC}$

4.0Detailed Step-by-Step Proof of BPT

To prove this theorem rigorously, we use a classic textbook method based on calculating and comparing the areas of specific triangles within the figure.

1. Given Data & Goal

Given: A triangle ABC where line DE is parallel to BC $(D E ∥ BC)$ , intersecting AB at D and AC at E.
To Prove: $\frac{A D}{D B} = \frac{A E}{EC}$

2. Construction Lines Required

To set up our area calculations, we need to add a few construction lines to the triangle:

Join vertex B to point E to form triangle BDE.
Join vertex C to point D to form triangle CDE.
Draw a perpendicular altitude from point E to the side AB, labelling it EN $(EN ⊥ A B)$ .
Draw a perpendicular altitude from point D to the side AC, labelling it DM $(D M ⊥ A C)$ .

3. Derivation Steps

Recall that the standard formula for the area of any triangle is:

$Area = \frac{1}{2} \times Base \times Height$

Let us analyze the left side of the triangle using the perpendicular height $EN$:

Area of $Δ A D E$ (taking AD as the base): $Area (A D E) = \frac{1}{2} \times A D \times EN$
Area of $Δ B D E$ (taking DB as the base): $Area (B D E) = \frac{1}{2} \times D B \times EN$

Now, let us find the ratio of these two areas:

$\frac{Area ( A D E )}{Area ( B D E )} = \frac{\frac{1}{2} \times A D \times EN}{\frac{1}{2} \times D B \times EN} = \frac{A D}{D B} — (Equation 1)$

Next, let us analyze the right side of the triangle using the perpendicular height $DM$:

Area of \Delta ADE$ (taking AE as the base):
$Area (A D E) = \frac{1}{2} \times A E \times D M$
Area of \Delta CDE (taking $EC$ as the base):
$Area (C D E) = \frac{1}{2} \times EC \times D M$

Let us find the ratio of these two areas as well:

$\frac{Area ( A D E )}{Area ( C D E )} = \frac{\frac{1}{2} \times A E \times D M}{\frac{1}{2} \times EC \times D M} = \frac{A E}{EC} — (Equation 2)$

4. The Critical Property

Look closely at the denominators of our two ratio equations: BDE and CDE

Both of these triangles sit on the exact same base line (DE).
Both triangles are nestled between the same pair of parallel lines (DE and BC).

According to a fundamental rule of geometry, triangles on the same base and between the same parallels are equal in area. Therefore:

$Area (B D E) = Area (C D E)$

Since the numerators in Equation 1 and Equation 2 are identical $(Area (A D E))$ and their denominators are equal, the entire left-hand side of both equations matches perfectly. Equating the right-hand sides gives us:

$\frac{A D}{D B} = \frac{A E}{EC}$

5.0What is the Basic Proportionality Theorem (BPT)?

The Basic Proportionality Theorem states: If a line is drawn parallel to one side of a triangle and intersects the other two sides at distinct points, then it divides those two sides in the same ratio.

Consider a triangle ABC.

Let a line DE be drawn parallel to BC such that:

D lies on AB
E lies on AC
DE ∥ BC

Then, according to the theorem:

$\frac{A D}{D B} = \frac{A E}{EC}$

This means that the line parallel to one side divides the other two sides proportionally.

6.0Diagram Representation of BPT

In triangle ABC:

DE is parallel to BC
D lies on AB
E lies on AC

Then:

AD : DB = AE : EC

This proportional relationship is the essence of the Basic Proportionality Theorem.

7.0Statement of Basic Proportionality Theorem

Theorem: If a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points, then the other two sides are divided in the same ratio.

This theorem is also referred to as Thales' Theorem in elementary geometry.

8.0Proof of Basic Proportionality Theorem

Given:

In △ABC, DE ∥ BC
D lies on AB
E lies on AC

To Prove:

$\frac{A D}{D B} = \frac{A E}{EC}$

Proof

Since DE is parallel to BC,

Triangles BDE and CDE lie between the same parallels and have the same altitude.

Using the property of equal areas and proportional segments formed by parallel lines, we obtain:

AD/DB = AE/EC

Hence proved.

This proof is commonly included in the CBSE Class 10 syllabus and forms an important examination topic.

9.0Converse of Basic Proportionality Theorem

The converse of BPT is equally important.

Statement

If a line divides any two sides of a triangle in the same ratio, then the line is parallel to the third side.

Mathematically,

If $\frac{A D}{D B} = \frac{A E}{EC}$

Then, DE ∥ BC

The converse is frequently used in geometry proofs and construction problems.

10.0Formula of Basic Proportionality Theorem

The main BPT formula is:

$\frac{A D}{D B} = \frac{A E}{EC}$

Alternative forms include:

$\frac{A D}{A B} = \frac{A E}{A C}$

and

$\frac{D B}{A B} = \frac{EC}{A C}$

These forms are useful while solving numerical questions.

11.0Applications of Basic Proportionality Theorem

The theorem has numerous practical and mathematical applications.

1. Similar Triangles: BPT is used extensively in proving the similarity of triangles.

2. Geometric Constructions: It helps divide line segments into equal or proportional parts.

3. Surveying: Surveyors use proportional relationships to measure inaccessible distances.

4. Architecture and Engineering: Scaled drawings and structural designs often rely on proportionality principles.

12.0The Converse of Basic Proportionality Theorem

What happens if we reverse the logic of BPT? If we encounter a triangle where the side segments form matching ratios, can we conclude that the intersecting line is parallel to the base? Yes! This logical reverse is known as the Converse of BPT.

The Converse Statement:

If a line divides any two sides of a triangle in the same ratio, then the line must be parallel to the third side.

If you are evaluating a triangle ABC and find that a line DE cuts the sides such that $\frac{A D}{D B} = \frac{A E}{EC}$

, you can immediately state with absolute geometric certainty that DE || BC. This converse is incredibly helpful for verification problems.

13.0Step-by-Step Solved Examples

Let us look at three distinct algebraic examples to see how BPT and its converse are applied in practice.

Example 1: Finding a missing segment value

Question: In $\Delta ABC$, a line $DE \parallel BC$ cuts through the sides $AB$ and $AC$. If the segments are measured as $AD = 3\text{ cm}$, $DB = 4.5\text{ cm}$, and $AE = 2\text{ cm}$, calculate the exact length of segment $EC$.

Solution:

Step 1: Confirm the conditions for BPT are met. Because $DE \parallel BC$, we can safely apply the Basic Proportionality Theorem.
Step 2: Write down the standard BPT formula.
$$\frac{AD}{DB} = \frac{AE}{EC}$$
Step 3: Substitute the given numerical values into the proportion.
$$\frac{3}{4.5} = \frac{2}{EC}$$
Step 4: Cross-multiply to isolate and solve for the unknown variable $EC$.
$$3 \times EC = 2 \times 4.5$$
$$3 \times EC = 9$$
$$EC = \frac{9}{3} = 3\text{ cm}$$

Answer: The length of segment $EC$ is 3 cm.

Example 2: Solving for an unknown variable '$x$'

Question: In a triangle $ABC$, a line parallel to the base $BC$ intersects the sides at $D$ and $E$. The segment lengths are expressed algebraically as $AD = x$, $DB = x + 3$, $AE = x + 1$, and $EC = x + 5$. Solve for the value of $x$.

Solution:

Step 1: Apply the BPT equation to set up an algebraic expression.
$$\frac{x}{x + 3} = \frac{x + 1}{x + 5}$$
Step 2: Cross-multiply to remove the fractional denominators.
$$x(x + 5) = (x + 3)(x + 1)$$
Step 3: Expand both sides of the equation.
$$x^2 + 5x = x^2 + 1x + 3x + 3$$
$$x^2 + 5x = x^2 + 4x + 3$$
Step 4: Simplify by subtracting $x^2$ from both sides, then isolate $x$.
$$5x = 4x + 3$$
$$5x - 4x = 3 \implies x = 3$$

Answer: The value of $x$ is 3.

Example 3: Verifying parallelism using the Converse of BPT

Question: In a triangle $PQR$, points $S$ and $T$ lie on the sides $PQ$ and $PR$ respectively. The total length of $PQ = 15\text{ cm}$ with $PS = 6\text{ cm}$, and the total length of $PR = 25\text{ cm}$ with $PT = 10\text{ cm}$. Determine whether line segment $ST$ is parallel to the base $QR$.

Solution:

Step 1: Calculate the lengths of the remaining lower segments ($SQ$ and $TR$).
$$SQ = PQ - PS = 15 - 6 = 9\text{ cm}$$
$$TR = PR - PT = 25 - 10 = 15\text{ cm}$$
Step 2: Calculate the numerical ratio of the left-hand segments.
$$\frac{PS}{SQ} = \frac{6}{9} = \frac{2}{3}$$
Step 3: Calculate the numerical ratio of the right-hand segments.
$$\frac{PT}{TR} = \frac{10}{15} = \frac{2}{3}$$
Step 4: Compare the two resulting ratios.
Since $\frac{PS}{SQ} = \frac{PT}{TR} = \frac{2}{3}$, the line $ST$ cuts the sides in matching proportions. According to the Converse of BPT, this confirms parallelism.

Answer: Yes, $ST$ is parallel to $QR$ ($ST \parallel QR$).

14.0Difference Between BPT and Similarity Criteria

Basis	Basic Proportionality Theorem	Similarity Criteria
Purpose	Establishes proportional division of sides	Proves triangles are similar
Condition	Requires a parallel line	Requires angle or side relationships
Application	Ratios and segment lengths	Similar triangles
Class Level	Class 10 Geometry	Class 10 Geometry

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16.0Supporting Study Materials

This study material, including CBSE Notes and NCERT Solutions for the Chapter "Triangles" focusing on Thales' Theorem, is designed according to the latest CBSE Class 10 Mathematics syllabus and NCERT guidelines. It contains exact geometric line models, area ratio equations, and high-yield question templates to ensure structural mastery for school tests and board exams.

CBSE Class 10 Science Notes Chapter 8 Heredity	NCERT Solution Class 10 Science Chapter 8: Heredity

17.030-Second Quick Revision: Basic Proportionality Theorem (BPT)

Also known as Thales' Theorem.
If a line is drawn parallel to one side of a triangle, it divides the other two sides in the same ratio.
In ΔABC, if DE ∥ BC, then:
AD/DB = AE/EC
A line parallel to one side of a triangle divides the other two sides proportionally.
Used to prove the similarity of triangles.
Converse of BPT is also true.
Helps in finding unknown lengths in geometrical figures.
Applicable only when the line is parallel to one side of the triangle.
One of the most important theorems in Class 10 Geometry.

18.0Previous Year Question (CBSE Class 10) On BPT

Question: In ΔABC, D and E are points on AB and AC respectively such that DE ∥ BC. If AD = 3 cm, DB = 6 cm and AE = 4 cm, find EC.

Solution: Using BPT,

AD/DB = AE/EC

3/6 = 4/EC

1/2 = 4/EC

EC = 8 cm

Question: In ΔABC, a line DE parallel to BC intersects AB at D and AC at E. If AD = 4 cm, DB = 8 cm and AE = 5 cm, find AC.

Solution: Using BPT,

AD/DB = AE/EC

4/8 = 5/EC

1/2 = 5/EC

EC = 10 cm

Therefore,

AC = AE + EC

AC = 5 + 10

AC = 15 cm

Answer: AC = 15 cm

19.0Recommended Next Topics

Converse of Basic Proportionality Theorem
Similarity Criterion (AA, SAS, SSS)
Areas of Similar Triangles
Pythagoras Theorem
Tangents to a Circle

Basic Proportionality Theorem (BPT)

1.0Master Thales' Theorem in Minutes

Class: 10 Mathematics (CBSE)
Chapter: Triangles
Estimated Learning Time: 25–30 Minutes

2.0Learning Outcomes

After completing this lesson, you will be able to:

State the Basic Proportionality Theorem (BPT) or Thales' Theorem clearly.
Set up construction elements and derive the full geometric proof using triangle areas.
State and apply the Converse of BPT to verify parallel lines in geometric figures.
Identify alternate mathematical variations of the BPT ratio formula.
Solve algebraic linear and quadratic equations based on proportional side segments.

The mathematical backbone of his discovery is what we study today as the Basic Proportionality Theorem (BPT), or simply Thales's Theorem.

3.0The Statement of Basic Proportionality Theorem (BPT)

The theorem establishes a strict proportional relationship between the line segments created when a parallel line cuts through a triangle.

The BPT Statement:

If a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points, then the other two sides are divided in the same ratio.

To look at this mathematically, let us consider a triangle ABC:

Suppose a straight line DE is drawn such that it is strictly parallel to the base side BC $(D E ∥ BC)$
This line intersects the side AB at point D and the side AC at point E.

According to the Basic Proportionality Theorem, this structural arrangement guarantees that:

$\frac{A D}{D B} = \frac{A E}{EC}$

4.0Detailed Step-by-Step Proof of BPT

To prove this theorem rigorously, we use a classic textbook method based on calculating and comparing the areas of specific triangles within the figure.

1. Given Data & Goal

Given: A triangle ABC where line DE is parallel to BC $(D E ∥ BC)$ , intersecting AB at D and AC at E.
To Prove: $\frac{A D}{D B} = \frac{A E}{EC}$

2. Construction Lines Required

To set up our area calculations, we need to add a few construction lines to the triangle:

Join vertex B to point E to form triangle BDE.
Join vertex C to point D to form triangle CDE.
Draw a perpendicular altitude from point E to the side AB, labelling it EN $(EN ⊥ A B)$ .
Draw a perpendicular altitude from point D to the side AC, labelling it DM $(D M ⊥ A C)$ .

3. Derivation Steps

Recall that the standard formula for the area of any triangle is:

$Area = \frac{1}{2} \times Base \times Height$

Let us analyze the left side of the triangle using the perpendicular height $EN$:

Area of $Δ A D E$ (taking AD as the base): $Area (A D E) = \frac{1}{2} \times A D \times EN$
Area of $Δ B D E$ (taking DB as the base): $Area (B D E) = \frac{1}{2} \times D B \times EN$

Now, let us find the ratio of these two areas:

$\frac{Area ( A D E )}{Area ( B D E )} = \frac{\frac{1}{2} \times A D \times EN}{\frac{1}{2} \times D B \times EN} = \frac{A D}{D B} — (Equation 1)$

Next, let us analyze the right side of the triangle using the perpendicular height $DM$:

Area of \Delta ADE$ (taking AE as the base):
$Area (A D E) = \frac{1}{2} \times A E \times D M$
Area of \Delta CDE (taking $EC$ as the base):
$Area (C D E) = \frac{1}{2} \times EC \times D M$

Let us find the ratio of these two areas as well:

$\frac{Area ( A D E )}{Area ( C D E )} = \frac{\frac{1}{2} \times A E \times D M}{\frac{1}{2} \times EC \times D M} = \frac{A E}{EC} — (Equation 2)$

4. The Critical Property

Look closely at the denominators of our two ratio equations: BDE and CDE

Both of these triangles sit on the exact same base line (DE).
Both triangles are nestled between the same pair of parallel lines (DE and BC).

According to a fundamental rule of geometry, triangles on the same base and between the same parallels are equal in area. Therefore:

$Area (B D E) = Area (C D E)$

$\frac{A D}{D B} = \frac{A E}{EC}$

5.0What is the Basic Proportionality Theorem (BPT)?

Consider a triangle ABC.

Let a line DE be drawn parallel to BC such that:

D lies on AB
E lies on AC
DE ∥ BC

Then, according to the theorem:

$\frac{A D}{D B} = \frac{A E}{EC}$

This means that the line parallel to one side divides the other two sides proportionally.

6.0Diagram Representation of BPT

In triangle ABC:

DE is parallel to BC
D lies on AB
E lies on AC

Then:

AD : DB = AE : EC

This proportional relationship is the essence of the Basic Proportionality Theorem.

7.0Statement of Basic Proportionality Theorem

Theorem: If a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points, then the other two sides are divided in the same ratio.

This theorem is also referred to as Thales' Theorem in elementary geometry.

8.0Proof of Basic Proportionality Theorem

Given:

In △ABC, DE ∥ BC
D lies on AB
E lies on AC

To Prove:

$\frac{A D}{D B} = \frac{A E}{EC}$

Proof

Since DE is parallel to BC,

Triangles BDE and CDE lie between the same parallels and have the same altitude.

Using the property of equal areas and proportional segments formed by parallel lines, we obtain:

AD/DB = AE/EC

Hence proved.

This proof is commonly included in the CBSE Class 10 syllabus and forms an important examination topic.

9.0Converse of Basic Proportionality Theorem

The converse of BPT is equally important.

Statement

If a line divides any two sides of a triangle in the same ratio, then the line is parallel to the third side.

Mathematically,

If $\frac{A D}{D B} = \frac{A E}{EC}$

Then, DE ∥ BC

The converse is frequently used in geometry proofs and construction problems.

10.0Formula of Basic Proportionality Theorem

The main BPT formula is:

$\frac{A D}{D B} = \frac{A E}{EC}$

Alternative forms include:

$\frac{A D}{A B} = \frac{A E}{A C}$

and

$\frac{D B}{A B} = \frac{EC}{A C}$

These forms are useful while solving numerical questions.

11.0Applications of Basic Proportionality Theorem

The theorem has numerous practical and mathematical applications.

1. Similar Triangles: BPT is used extensively in proving the similarity of triangles.

2. Geometric Constructions: It helps divide line segments into equal or proportional parts.

3. Surveying: Surveyors use proportional relationships to measure inaccessible distances.

4. Architecture and Engineering: Scaled drawings and structural designs often rely on proportionality principles.

12.0The Converse of Basic Proportionality Theorem

The Converse Statement:

If a line divides any two sides of a triangle in the same ratio, then the line must be parallel to the third side.

If you are evaluating a triangle ABC and find that a line DE cuts the sides such that $\frac{A D}{D B} = \frac{A E}{EC}$

, you can immediately state with absolute geometric certainty that DE || BC. This converse is incredibly helpful for verification problems.

13.0Step-by-Step Solved Examples

Let us look at three distinct algebraic examples to see how BPT and its converse are applied in practice.

Example 1: Finding a missing segment value

Solution:

Step 1: Confirm the conditions for BPT are met. Because $DE \parallel BC$, we can safely apply the Basic Proportionality Theorem.
Step 2: Write down the standard BPT formula.
$$\frac{AD}{DB} = \frac{AE}{EC}$$
Step 3: Substitute the given numerical values into the proportion.
$$\frac{3}{4.5} = \frac{2}{EC}$$
Step 4: Cross-multiply to isolate and solve for the unknown variable $EC$.
$$3 \times EC = 2 \times 4.5$$
$$3 \times EC = 9$$
$$EC = \frac{9}{3} = 3\text{ cm}$$

Answer: The length of segment $EC$ is 3 cm.

Example 2: Solving for an unknown variable '$x$'

Solution:

Step 1: Apply the BPT equation to set up an algebraic expression.
$$\frac{x}{x + 3} = \frac{x + 1}{x + 5}$$
Step 2: Cross-multiply to remove the fractional denominators.
$$x(x + 5) = (x + 3)(x + 1)$$
Step 3: Expand both sides of the equation.
$$x^2 + 5x = x^2 + 1x + 3x + 3$$
$$x^2 + 5x = x^2 + 4x + 3$$
Step 4: Simplify by subtracting $x^2$ from both sides, then isolate $x$.
$$5x = 4x + 3$$
$$5x - 4x = 3 \implies x = 3$$

Answer: The value of $x$ is 3.

Example 3: Verifying parallelism using the Converse of BPT

Solution:

Step 1: Calculate the lengths of the remaining lower segments ($SQ$ and $TR$).
$$SQ = PQ - PS = 15 - 6 = 9\text{ cm}$$
$$TR = PR - PT = 25 - 10 = 15\text{ cm}$$
Step 2: Calculate the numerical ratio of the left-hand segments.
$$\frac{PS}{SQ} = \frac{6}{9} = \frac{2}{3}$$
Step 3: Calculate the numerical ratio of the right-hand segments.
$$\frac{PT}{TR} = \frac{10}{15} = \frac{2}{3}$$
Step 4: Compare the two resulting ratios.
Since $\frac{PS}{SQ} = \frac{PT}{TR} = \frac{2}{3}$, the line $ST$ cuts the sides in matching proportions. According to the Converse of BPT, this confirms parallelism.

Answer: Yes, $ST$ is parallel to $QR$ ($ST \parallel QR$).

14.0Difference Between BPT and Similarity Criteria

Basis	Basic Proportionality Theorem	Similarity Criteria
Purpose	Establishes proportional division of sides	Proves triangles are similar
Condition	Requires a parallel line	Requires angle or side relationships
Application	Ratios and segment lengths	Similar triangles
Class Level	Class 10 Geometry	Class 10 Geometry

15.0EUREKA by ALLEN – The Future of Class 10 Learning

Key Features of EUREKA Class 10 Courses:
Easy-to-follow concept explanations Story-based learning for better retention Exam-oriented preparation strategy Answer writing practice sessions Instant quizzes and self-assessment Performance tracking dashboard Round-the-clock AI doubt support NCERT-based curriculum coverage Study at your own pace

Explore Now

16.0Supporting Study Materials

CBSE Class 10 Science Notes Chapter 8 Heredity	NCERT Solution Class 10 Science Chapter 8: Heredity

17.030-Second Quick Revision: Basic Proportionality Theorem (BPT)

Also known as Thales' Theorem.
If a line is drawn parallel to one side of a triangle, it divides the other two sides in the same ratio.
In ΔABC, if DE ∥ BC, then:
AD/DB = AE/EC
A line parallel to one side of a triangle divides the other two sides proportionally.
Used to prove the similarity of triangles.
Converse of BPT is also true.
Helps in finding unknown lengths in geometrical figures.
Applicable only when the line is parallel to one side of the triangle.
One of the most important theorems in Class 10 Geometry.

18.0Previous Year Question (CBSE Class 10) On BPT

Question: In ΔABC, D and E are points on AB and AC respectively such that DE ∥ BC. If AD = 3 cm, DB = 6 cm and AE = 4 cm, find EC.

Solution: Using BPT,

AD/DB = AE/EC

3/6 = 4/EC

1/2 = 4/EC

EC = 8 cm

Question: In ΔABC, a line DE parallel to BC intersects AB at D and AC at E. If AD = 4 cm, DB = 8 cm and AE = 5 cm, find AC.

Solution: Using BPT,

AD/DB = AE/EC

4/8 = 5/EC

1/2 = 5/EC

EC = 10 cm

Therefore,

AC = AE + EC

AC = 5 + 10

AC = 15 cm

Answer: AC = 15 cm

19.0Recommended Next Topics

Converse of Basic Proportionality Theorem
Similarity Criterion (AA, SAS, SSS)
Areas of Similar Triangles
Pythagoras Theorem
Tangents to a Circle

Frequently Asked Questions

What is the Basic Proportionality Theorem?

Why is BPT called Thales' Theorem?

What is the formula of BPT?

What is the converse of the Basic Proportionality Theorem?

Where is the Basic Proportionality Theorem used?

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Basic Proportionality Theorem (BPT)

1.0Master Thales' Theorem in Minutes

2.0Learning Outcomes

3.0The Statement of Basic Proportionality Theorem (BPT)

The BPT Statement:

4.0Detailed Step-by-Step Proof of BPT

1. Given Data & Goal

2. Construction Lines Required

3. Derivation Steps

4. The Critical Property

5.0What is the Basic Proportionality Theorem (BPT)?

6.0Diagram Representation of BPT

7.0Statement of Basic Proportionality Theorem

8.0Proof of Basic Proportionality Theorem

Proof

9.0Converse of Basic Proportionality Theorem

Statement

10.0Formula of Basic Proportionality Theorem

11.0Applications of Basic Proportionality Theorem

12.0The Converse of Basic Proportionality Theorem

The Converse Statement:

13.0Step-by-Step Solved Examples

Example 1: Finding a missing segment value

Example 2: Solving for an unknown variable '$x$'

Example 3: Verifying parallelism using the Converse of BPT

14.0Difference Between BPT and Similarity Criteria

15.0EUREKA by ALLEN – The Future of Class 10 Learning

16.0Supporting Study Materials

17.030-Second Quick Revision: Basic Proportionality Theorem (BPT)

18.0Previous Year Question (CBSE Class 10) On BPT

19.0Recommended Next Topics

Table of Contents

Frequently Asked Questions

What is the Basic Proportionality Theorem?

Why is BPT called Thales' Theorem?

What is the formula of BPT?

What is the converse of the Basic Proportionality Theorem?

Where is the Basic Proportionality Theorem used?

Join ALLEN!

Join ALLEN!

Basic Proportionality Theorem (BPT)

1.0Master Thales' Theorem in Minutes

2.0Learning Outcomes

3.0The Statement of Basic Proportionality Theorem (BPT)

The BPT Statement:

4.0Detailed Step-by-Step Proof of BPT

1. Given Data & Goal

2. Construction Lines Required

3. Derivation Steps

4. The Critical Property

5.0What is the Basic Proportionality Theorem (BPT)?

6.0Diagram Representation of BPT

7.0Statement of Basic Proportionality Theorem

8.0Proof of Basic Proportionality Theorem

Proof

9.0Converse of Basic Proportionality Theorem

Statement

10.0Formula of Basic Proportionality Theorem

11.0Applications of Basic Proportionality Theorem

12.0The Converse of Basic Proportionality Theorem

The Converse Statement:

13.0Step-by-Step Solved Examples

Example 1: Finding a missing segment value

Example 2: Solving for an unknown variable '$x$'

Example 3: Verifying parallelism using the Converse of BPT

14.0Difference Between BPT and Similarity Criteria

15.0EUREKA by ALLEN – The Future of Class 10 Learning

16.0Supporting Study Materials

17.030-Second Quick Revision: Basic Proportionality Theorem (BPT)

18.0Previous Year Question (CBSE Class 10) On BPT

19.0Recommended Next Topics

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