Basic Proportionality Theorem (BPT): Statement, Proof, Formula, Examples & Applications

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Basic Proportionality Theorem

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1.0Basic Proportionality Theorem (BPT)

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.

2.0The 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:

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

3.0Detailed 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):
$$\text{Area}(ADE) = \frac{1}{2} \times AE \times DM$$
Area of $\Delta CDE$ (taking $EC$ as the base):
$$\text{Area}(CDE) = \frac{1}{2} \times EC \times DM$$

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

$$\frac{\text{Area}(ADE)}{\text{Area}(CDE)} = \frac{\frac{1}{2} \times AE \times DM}{\frac{1}{2} \times EC \times DM} = \frac{AE}{EC} \quad \text{--- (Equation 2)}$$

4. The Critical Property

Look closely at the denominators of our two ratio equations: $\Delta BDE$ and $\Delta CDE$.

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:

$$\text{Area}(BDE) = \text{Area}(CDE)$$

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

$$\frac{AD}{DB} = \frac{AE}{EC}$$

4.0What 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{AD}{DB}=\frac{AE}{EC}

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

5.0Diagram 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.

6.0Statement 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.

7.0Proof of Basic Proportionality Theorem

Given:

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

To Prove:

\frac{AD}{DB}=\frac{AE}{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.

8.0Converse 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{AD}{DB}=\frac{AE}{EC}

Then,

DE ∥ BC

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

9.0Formula of Basic Proportionality Theorem

The main BPT formula is:

\frac{AD}{DB}=\frac{AE}{EC}

Alternative forms include:

\frac{AD}{AB}=\frac{AE}{AC}

and

\frac{DB}{AB}=\frac{EC}{AC}

These forms are useful while solving numerical questions.

10.0Applications 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.

11.0The 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 $\frac{AD}{DB} = \frac{AE}{EC}$, you can immediately state with absolute geometric certainty that $DE \parallel BC$. This converse is incredibly helpful for verification problems.

12.0Step-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 $QR$ ($ST \parallel QR$).

13.0Difference 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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1.0Basic Proportionality Theorem (BPT)

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.

2.0The 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:

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

3.0Detailed 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):
$$\text{Area}(ADE) = \frac{1}{2} \times AE \times DM$$
Area of $\Delta CDE$ (taking $EC$ as the base):
$$\text{Area}(CDE) = \frac{1}{2} \times EC \times DM$$

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

$$\frac{\text{Area}(ADE)}{\text{Area}(CDE)} = \frac{\frac{1}{2} \times AE \times DM}{\frac{1}{2} \times EC \times DM} = \frac{AE}{EC} \quad \text{--- (Equation 2)}$$

4. The Critical Property

Look closely at the denominators of our two ratio equations: $\Delta BDE$ and $\Delta CDE$.

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:

$$\text{Area}(BDE) = \text{Area}(CDE)$$

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

$$\frac{AD}{DB} = \frac{AE}{EC}$$

4.0What 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{AD}{DB}=\frac{AE}{EC}

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

5.0Diagram 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.

6.0Statement 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.

7.0Proof of Basic Proportionality Theorem

Given:

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

To Prove:

\frac{AD}{DB}=\frac{AE}{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.

8.0Converse 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{AD}{DB}=\frac{AE}{EC}

Then,

DE ∥ BC

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

9.0Formula of Basic Proportionality Theorem

The main BPT formula is:

\frac{AD}{DB}=\frac{AE}{EC}

Alternative forms include:

\frac{AD}{AB}=\frac{AE}{AC}

and

\frac{DB}{AB}=\frac{EC}{AC}

These forms are useful while solving numerical questions.

10.0Applications 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.

11.0The 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 $\frac{AD}{DB} = \frac{AE}{EC}$, you can immediately state with absolute geometric certainty that $DE \parallel BC$. This converse is incredibly helpful for verification problems.

12.0Step-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 $QR$ ($ST \parallel QR$).

13.0Difference 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

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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1.0Basic Proportionality Theorem (BPT)

2.0The Statement of Basic Proportionality Theorem (BPT)

The BPT Statement:

3.0Detailed Step-by-Step Proof of BPT

1. Given Data & Goal

2. Construction Lines Required

3. Derivation Steps

4. The Critical Property

4.0What is the Basic Proportionality Theorem (BPT)?

5.0Diagram Representation of BPT

6.0Statement of Basic Proportionality Theorem

7.0Proof of Basic Proportionality Theorem

Proof

8.0Converse of Basic Proportionality Theorem

Statement

9.0Formula of Basic Proportionality Theorem

10.0Applications of Basic Proportionality Theorem

11.0The Converse of Basic Proportionality Theorem

The Converse Statement:

12.0Step-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

13.0Difference Between BPT and Similarity Criteria

On this page

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!

1.0Basic Proportionality Theorem (BPT)

2.0The Statement of Basic Proportionality Theorem (BPT)

The BPT Statement:

3.0Detailed Step-by-Step Proof of BPT

1. Given Data & Goal

2. Construction Lines Required

3. Derivation Steps

4. The Critical Property

4.0What is the Basic Proportionality Theorem (BPT)?

5.0Diagram Representation of BPT

6.0Statement of Basic Proportionality Theorem

7.0Proof of Basic Proportionality Theorem

Proof

8.0Converse of Basic Proportionality Theorem

Statement

9.0Formula of Basic Proportionality Theorem

10.0Applications of Basic Proportionality Theorem

11.0The Converse of Basic Proportionality Theorem

The Converse Statement:

12.0Step-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

13.0Difference Between BPT and Similarity Criteria

On this page