The Basic Proportionality Theorem was proposed by Thales, a Greek mathematician. It gives the relationship between the sides of any two equiangular triangles. The theorem also helps us better understand the concept of similar triangles.

Let us start this lesson with a simple riddle. Identify the number of triangles in the below figure.

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Now let us try and understand the Basic Proportionality Theorem.

**Lesson Plan**

**What Is the Statement for Basic Proportionality Theorem?**

The line drawn parallel to one side of a triangle and cutting the other two sides divides the other two sides in equal proportion.

Here DE is parallel to BC. \[\dfrac{AD}{DB} = \dfrac{AE}{EC} \]

The Basic Proportionality Theorem is also referred to as Thales Theorem

**How Does One Prove the Basic Proportionality Theorem? **

In the above diagram, create imaginary lines where you can Join C to D and B to E

Draw perpendicular DP perpendicular to AE and EQ perpendicular to AD

**Proof:**

Consider the triangles ADE and BDE, taken from the triangle ABC.

Both these triangles are on the same base AB and have equal height EQ.

\[ \begin{align} \dfrac{Area~of~ADE}{Area~of~BDE} & = \dfrac{\frac{1}{2} \times AD \times EQ}{ \frac{1}{2} \times BD \times EQ} \\ \dfrac{Area~of~ADE}{Area~of~BDE} &=\dfrac{AD}{BD}\end{align}\]

Now consider triangles CDE and ADE.

Both these triangles are on the same base AC and have equal height DP.

\[ \begin{align} \dfrac{Area~of~ADE}{Area~of~CDE} &= \dfrac{\frac{1}{2} \times AE \times DP}{ \frac{1}{2} \times CE \times DP } \\ \dfrac{Area~of~ADE}{Area~of~CDE} &= \dfrac{AE}{CE}\end{align}\]

Both the triangles BDE and CDE are between the same set of parallel lines.

\[ Area~of~traingle~BDE = Area~of~traingle~CDE\]

Applying this we have:

\[ \begin{align}\dfrac{Area~of~traingle~ADE}{ Area~of~traingle~BDE} &= \dfrac{Area~of~traingle~ADE}{ Area~of~traingle~CDE} \\ \dfrac{AD}{BD}&= \dfrac{AE}{CE}\end{align} \]

**Corollary:**

The above proof is also helpful to prove another important theorem called the mid-point theorem.

The mid-point theorem states that a line segment drawn parallel to one side of a triangle and half of that side divides the other two sides at the midpoints.

**Conclusion:**

Hence we prove the Basic Proportionality Theorem.

Therefore the line DE drawn parallel to the side BC of triangle ABC divides the other two sides AB, AC in equal proportion.

- Basic Proportionality Theorem - A line drawn parallel to one side of a triangle and cutting the other two sides, divides the other two sides in equal proportion.
- Converse of Basic Proportionality Theorem - A line drawn to cut two sides of a triangle in equal proportion is parallel to the third side.
- Midpoint Theorem - A line drawn parallel to one side of the triangle and half of that side, divides the other two sides at its midpoint.

**What Is the Converse of Basic Proportionality Theorem?**

A line segment is drawn to cut two sides of a triangle in equal proportion, then it is parallel to the third side.

**Given:**

ABC is a triangle and the line DE cuts the sides AB and AC in equal proportion. \[\dfrac{AD}{BD}= \dfrac{AE}{CE} \]

**Proof:**

Consider DE is not parallel to BC. Hence let us draw another line DF which is parallel to BC.

Applying the Basic Proportionality theorem we have: \[\dfrac{AD}{BD}= \dfrac{AF}{CF} \]

But it is already given that : \[\dfrac{AD}{BD}= \dfrac{AE}{CE} \]

Observing the right-hand sides of the above two statements we have the following statement. \[\dfrac{AE}{CE}= \dfrac{AF}{CF} \]

Add 1 on both sides of this statement.

\[\begin{align} \dfrac{AE}{CE} + 1&= \dfrac{AF}{CF} + 1\\ \dfrac{AE + CE}{CE}&= \dfrac{AF + CF}{CF} \\ \dfrac{AC}{CE}&= \dfrac{AC}{CF} \\ \therefore CE & = CF\end{align} \]

For the above statement, the points E and F are the same points and they are coincident.

Hence the line DE is parallel to BC and it proves the converse of the basic proportionality theorem.

**Solved Examples**

Example 1 |

In a triangle ABC, D is a point on the side AB and E is a point on the side AC. And the lengths AB = 10 inches, AC = 12 inches, AD = 5 inches, AE = 6 inches. Is the line DE parallel to BC.

**Solution**

For the traingle ABC we have AB = 10 inches, AC = 12 inches, AD = 5 inches, AE = 6 inches

BD = AB - AD = 10 - 5 = 5 inches

CE = AC - AE = 12 - 6 = 6 inches

\[\begin{align} \dfrac{AD}{BD} &= \dfrac{5}{5} = 1\\ \dfrac{AE}{CE} &= \dfrac{6}{6} = 1 \\ \dfrac{AD}{BD} &= \dfrac{AE}{CE} \end{align}\]

Applying the converse of basic proportionality theorem, the line segment DE is parallel to the side BC of the triangle ABC.

\(\therefore \) DE is parallel to BC |

Example 2 |

In the below figure DE// BC and EF // CD. Prove that \(\dfrac{AD}{BD} = \dfrac{AF}{DF} \)

**Solution**

In triangle ABC we have DE// BC and hence we have: \[ \dfrac{AD}{BD} = \dfrac{AE}{CE}\]

Similary in triangle ACD we have EF// CD. Here applying the basic proportionality theorem we have: \[ \dfrac{AE}{CE} = \dfrac{AF}{DF}\]

From the above two statements we can make the following conclusion. \[ \dfrac{AD}{BD} = \dfrac{AF}{DF}\]

\(\therefore \dfrac{AD}{BD} = \dfrac{AF}{DF} \) |

Example 3 |

In a triangle ABC \( \dfrac{AD}{BD} = \frac{AE}{CE}\) and \(\angle D = \angle C\) . Prove that ABC is an isocelous triangle.

**Solution**

It is given that \( \dfrac{AD}{BD} = \frac{AE}{CE}\)

Applying the converse of the basic proportionality theorem we can conclude that DE \\ BC.

Hence the corresponding angles are equal.\[\angle D = \angle B \]

Also, we are given. \[\angle D = \angle C \]

From the above two statements, we can conclude that the two corresponding angles are equal. \[\angle B = \angle C \]

And hence we have AB = AC

Therefore ABC is an isosceles triangle

\(\therefore \) ABC is an isosceles triangle. |

The diagonals of a quadrilateral PQRS intersect at the point O, such that \( \dfrac{PO}{QO} = \dfrac{RO}{SO} \).

Prove that PQRS is a trapezium.

**Interactive Questions on Basic Proportionality Theorem**

**Here are a few activities for you to practice. Select/Type your answer and click the "Check Answer" button to see the result.**

**Let's Summarize**

The mini-lesson targeted the fascinating concept of the basic proportionality theorem. The math journey around basic proportionality theorem starts with what a student already knows, and goes on to creatively crafting a fresh concept in the young minds. Done in a way that not only it is relatable and easy to grasp, but also will stay with them forever. Here lies the magic with Cuemath.

**About Cuemath**

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## Frequently Asked Questions(FAQs)

### 1. What is Thales theorem?

The Thales theorem, which is also referred to as the basic proportionality theorem, states that the line drawn parallel to one side and of a triangle and cutting the other two sides divides those two sides in equal proportion.

### 2. What are the applications of the basic proportionality theorem?

The basic proportionality theorem helps to find the lengths in which the two sides of a triangle are divided by a line drawn parallel to the third side. Further, it has applications to find the relationship between two equiangular triangles.

### 3. What is the history of Thales theorem?

The Thales Theorem was proposed by Thales, a Greek mathematician, and philosopher around 625 BC. It is now referred to as the basic proportionality theorem and it helps to find the relationship between the sides of two equiangular triangles.

### 4. What is the formula of the basic proportionality theorem?

The basis proportionality theorem formula for a triangle ABC with point D on AB, point E on AC, and DE // BC, is as follows.

\[ \dfrac{AD}{DB} = \dfrac{AE}{EC}\]

### 5. What do you mean by the basic proportionality theorem?

The basic proportionality theorem states that if a line is drawn parallel to one side of a triangle and is cutting the other two sides, then it divides the other two sides in equal proportion.

### 6. How do you prove the basic proportionality theorem by paper cutting?

To show the basic proportionality theorem, cut a colored paper in the shape of a triangle, and mark its vertices as ABC. Place it on a ruled paper with one side BC coinciding with a line on the ruled paper. Now mark points D on AB and E on AC such that DE is parallel to side BC. Now measure lengths AD, BD, AE, and CE and check if they are in proportion.

\[ \dfrac{AD}{DB} = \dfrac{AE}{EC}\]

### 7. How do you solve the Thales theorem?

The Thales theorem is the same as the basic proportionality theorem. To solve it we have to prove that the line drawn parallel to one side of the triangle, divides the other two sides in equal proportion.