Basic Proportionality Theorem
Basic proportionality theorem was proposed by a famous Greek mathematician, Thales, hence, it is also referred to as the Thales theorem. According to the famous mathematician, for any two equiangular triangles, the ratio of any two corresponding sides of the given triangles is always the same. Based on this concept, the basic proportionality theorem(BPT) was proposed. It gives the relationship between the sides of any two equiangular triangles.
The concept of Thales theorem has been introduced in similar triangles. If the given two triangles are similar to each other then,
 Corresponding angles of both the triangles are equal
 Corresponding sides of both the triangles are in proportion to each other
The theorem thus also helps us better understand the concept of similar triangles. Now let us try and understand the Basic Proportionality Theorem.
1.  Statement of Basic Proportionality Theorem 
2.  Proof of Basic Proportionality Theorem 
3.  Converse of Basic Proportionality Theorem 
4.  FAQs 
Statement of Basic Proportionality Theorem
The basic proportionality theorem, also known as the Thales theorem states that "the line drawn parallel to one side of a triangle and cutting the other two sides divides the other two sides in equal proportion". For example, in the given figure, line DE is drawn parallel to side BC, such that it joins the other two sides, AB and AC. According to the basic proportionality theorem, it can be implied that AD/DB = AE/EC.
Proof of the Basic Proportionality Theorem
Let us now try to prove the basic proportionality(BPT) theorem statement.
Statement: The line drawn parallel to one side of a triangle and cutting the other two sides divides the other two sides in equal proportion.
Given: Consider a triangle ΔABC, as shown in the given figure. In this triangle, we draw a line DE parallel to the side BC of ΔABC and intersecting the sides AB and AC at D and E, respectively.
Construction: 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. Both these triangles are on the same base AB and have equal height EQ.
(Area of ADE)/(Area of BDE) = (1/2 × AD × EQ)/(1/2 × BD × EQ)
(Area of ADE)/(Area of BDE) = AD/BD
Now consider triangles CDE and ADE. Both these triangles are on the same base AC and have equal height DP.
(Area of ADE)/(Area of CDE) = (1/2 × AE × DP)/(1/2 × CE × DP)
(Area of ADE)/(Area of CDE) = AE/CE
Both the triangles BDE and CDE are between the same set of parallel lines.
Area of triangle BDE = Area of triangle CDE
Applying this we have, (Area of triangle ADE)/(Area of triangle BDE) = (Area of triangle ADE)/(Area of triangle CDE)
AD/BD = AE/CE
Corollary:
The above proof is also helpful to prove another important theorem called the midpoint theorem. The midpoint 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. Also, the converse of the BPT, midpoint theorem also stands true. It states that the line drawn through the midpoint of a side of a triangle that is parallel to another side bisects the third side of the triangle.
Converse of Basic Proportionality Theorem
According to the converse of basic proportionality theorem, "If 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. AD/BD = 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: AD/BD = AF/CF. But it is already given that: AD/BD = AE/CE. Observing the equal lefthand sides of the above two statements we conclude the following statement, AE/CE = AF/CF. Add 1 on both sides of this statement.
(AE/CE) + 1= (AF/CF) + 1
(AE + CE)/CE = (AF + CF)/CF
AC/CE = AC/CF
∴ CE = CF
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.
Important Notes
 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.
 The 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.
Challenging Questions
 The diagonals of a quadrilateral PQRS intersect at the point O, such that PO/QO = RO/SO. Prove that PQRS is a trapezium.
Solved Examples on Basic Proportionality Theorem

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
So, AD/BD = 5/5 = 1 and AE/CE = 6/6 = 1. That means AD/BD = AE/CE. Applying the converse of basic proportionality theorem, the line segment DE is parallel to the side BC of the triangle ABC.
∴ DE is parallel to BC

Example 2: In the below figure GE//BC and EF//CD. Use basic proportionality theorem to prove AG/BG = AF/DF.
Solution:
In triangle ABC we have GE// BC and hence we have: AG/BG = AE/CE. Similarly in triangle ACD we have EF// CD. Here applying the basic proportionality theorem we have: AE/CE = AF/DF
From the above two statements, we can make the following conclusion. AG/BG = AF/DF
∴AG/BG = AF/DF

Example 3: In a triangle ABC, AD/BD = AE/CE and ∠ADE = ∠ACB. Prove that ABC is an isosceles triangle.
Solution:
It is given that AD/BD = AE/CE. Applying the converse of the basic proportionality theorem we can conclude that DE\\BC. Hence the corresponding angles are equal, ∠D = ∠B. Also, we are given, ∠D = ∠C
From the above two statements, we can conclude that the two corresponding angles are equal, ∠B = ∠C. And hence we have AB = AC. Therefore ABC is an isosceles triangle
Practice Questions on Basic Proportionality Theorem
FAQs on Basic Proportionality Theorem
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 of a triangle and cutting the other two sides divides those two sides in equal proportion.
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.
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.
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,
AD/DB = AE/EC
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.
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.
AD/DB = AE/EC
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.