The Basic Proportionality Theorem was developed by “Thales,” a prominent Greek mathematician. The Thales theorem is another name for this theory. “A line drawn parallel to one side of a triangle and cutting the other two sides splits the other two sides in equal proportion,” says the basic proportionality theorem, often known as the Thales theorem. The lengths of the intercepts made by a line segment on the other two sides of a triangle when it is drawn parallel to the third side of the triangle are described by this theorem. It is an important theorem in geometry that deals with triangles.

We recommend that students pay great attention to this theorem since it will help them in understanding basic mathematical concepts such as triangles in geometry. The more they understand, the easier it will be for them to achieve excellent exam scores.

Definition of Basic Proportionality Theorem

The basic proportionality theorem was found by a famous Greek mathematician Thales. It is also called the Thales theorem. 

The Basic proportionality Theorem or Thales Theorem states that if a line is drawn parallel to one side of a triangle and intersect the other two sides, then the line divides the two sides in the same ratio.

Let us consider a triangle \(ABC,\) such that line \(DE\) has drawn parallel to the base of the triangle \(BC.\) Then, the line \(DE\) divides the sides \(AB, AC\) in the same ratio.So, from the Basic proportionality Theorem or Thales Theorem, we get  \(\frac{{AD}}{{DB}} = \frac{{AE}}{{EC}}\)

Proof of Basic Proportionality Theorem

Let us look at the proof of the Basic proportionality theorem:

Statement:
In a triangle, if the line drawn parallel to one side of a triangle intersects the other sides at the two points, and then it divides the other two sides in an equal ratio.

Given:
Consider the triangle \(ABC\) shown in the figure below. In the given triangle, a line \(DE\) is drawn parallel to the base \(BC\) to intersecting sides \(AB, AC,\) of \(∆ABC\) at points \(D, E.\)

Construction:
Join \(BE\) and \(CD.\) Draw perpendicular lines \(DM\) and \(EN\) to the sides \(AC\) and \(AB,\) of \(∆ABC,\) respectively.

Proof:

We know that the area of a triangle is given by  \(\frac{1}{2} \times {\rm{base}} \times {\rm{height}}{\rm{.}}\)

Consider \(∆ADE,\) with the base as \(AD\) and height \(EN.\)

So, the area of rectangle \(∆ADE\) is \(\frac{1}{2} \times AD \times EN……..\left( 1 \right)\)

The area of \(∆BDE\) is \(\frac{1}{2} \times BD \times EN……..\left( 2 \right)\)

From the equations \(1, (2)\)

\(\frac{{{\mathop{\rm ar}\nolimits} (\Delta ADE)}}{{{\mathop{\rm ar}\nolimits} (\Delta BDE)}} = \frac{{\frac{1}{2} \times AD \times EN}}{{\frac{1}{2} \times BD \times EN}}\)

\( \Rightarrow \frac{{{\mathop{\rm ar}\nolimits} (\Delta ADE)}}{{{\mathop{\rm ar}\nolimits} (\Delta BDE)}} = \frac{{\frac{1}{2} \times AD \times EN}}{{\frac{1}{2} \times BD \times EN}}\)

\( \Rightarrow \frac{{{\mathop{\rm ar}\nolimits} (\Delta ADE)}}{{ar(\Delta BDE)}} = \frac{{AD}}{{BD}}………….\left( 3 \right)\)

Consider the triangle \(ADE,\) with base \(AE\) and height \(DM.\)

So, the area of rectangle ∆ADE is \(\frac{1}{2} \times AE \times DM……..\left( 4 \right)\)

The area of ∆CDE is \(\frac{1}{2} \times CE \times DM……..\left( 5 \right)\)

From the equations \(4, (5)\)

\(\frac{{{\mathop{\rm ar}\nolimits} (\Delta ADE)}}{{{\mathop{\rm ar}\nolimits} (\Delta CDE)}} = \frac{{\frac{1}{2} \times AE \times DM}}{{\frac{1}{2} \times CE \times DM}}\)

\( \Rightarrow \frac{{{\mathop{\rm ar}\nolimits} (\Delta ADE)}}{{{\mathop{\rm ar}\nolimits} (\Delta CDE)}} = \frac{{AE}}{{CE}}………\left( 6 \right)\)

We know that the area of triangles, which are having the same base and lying between the same parallel lines, are equal.

Here, \(∆BDE\) and \(∆CDE\) have the same base \(DE\) and lie in between the parallel lines \(DE\) and \(BC\left( {DC\parallel BC} \right).\)
So, their areas are equal.

\({\mathop{\rm ar}\nolimits} (\Delta BDE) = {\mathop{\rm ar}\nolimits} (\Delta CDE)……….{\rm{(7)}}\)

From equations \(3, 6\) and \((7),\) we get,  \(\frac{{{\mathop{\rm ar}\nolimits} (\Delta ADE)}}{{{\mathop{\rm ar}\nolimits} (\Delta BDE)}} = \frac{{{\mathop{\rm ar}\nolimits} (\Delta ADE)}}{{{\mathop{\rm ar}\nolimits} (\Delta CDE)}}.\)

So, we have

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

Hence, the Basic proportionality theorem is proved.

The Converse of Basic Proportionality Theorem Statement

This the reverse of the Basic Proportionality Theorem. The Converse of Basic Proportionality Theorem states that if a line divides any two sides of a triangle in the same ratio, then the line must be parallel to the third side of the triangle.

Let us consider a triangle \(ABC,\) in which line \(DE\) dividing the sides \(AB, AC\) in an equal ratio \(\left( {\frac{{AD}}{{DB}} = \frac{{AE}}{{EC}}} \right),\) then the line \(DE\) is parallel to the third side \(BC.\)

\(DE\parallel BC\)

The Converse of Basic Proportionality Theorem Proof

The proof of the Converse of Basic Proportionality Theorem is given below:

Given:
Consider a triangle \(ABC,\) in which \(DE\) intersects the sides \(AB, AC\) at \(D\) and \(E,\) such that  \(\frac{{AD}}{{DB}} = \frac{{AE}}{{EC}}.\)

Construction:
Let’s consider that \(DE\) is not parallel to \(BC.\) Then, there must be another line that is parallel to \(BC.\) Let this line be \(DE’.\) (Drawn by us). We shall now prove that \(DE\) and \(DE’\) are the same line or the points \(E\) and \(E’\) coincide.

Proof:

Given, \(\frac{{AD}}{{DB}} = \frac{{AE}}{{EC}}…………\left( 1 \right)\)

Here, \(DE’\parallel BC\) (drawn).
By basic proportionality theorem, we know that a line drawn parallel to one side divides the other sides in an equal ratio.

So, \(\frac{{AD}}{{DB}} = \frac{{AE’}}{{CE’}}….\left( 2 \right)\)

From \((1)\) and \((2,)\) we get, 

\(\frac{{AC}}{{EC}} = \frac{{AE’}}{{CE’}}\)

\( \Rightarrow \frac{{AE}}{{EC}} + 1 = \frac{{AE’}}{{CE’}} + 1\)

\( \Rightarrow \frac{{AE + EC}}{{EC}} = \frac{{AE’ + CE’}}{{CE’}}\)

From the above figure, we get, 

\(\frac{{AC}}{{EC}} = \frac{{AC}}{{C{E^\prime }}}\)

\( \Rightarrow \frac{1}{{EC}} = \frac{1}{{C{E^\prime }}}\)

\( \Rightarrow EC = C{E^\prime }………..\left( 3 \right)\)

It is possible only in the situation when \(E\) and \(E’\) coincide.
So, \(E=E’\) and thus, the lines \(DE\) and \(DE’\) are the same line.

Hence, the line \(DE\) is parallel to side \(BC.\)

Therefore, the Converse of Basic Proportionality Theorem is proved.

The Special Cases of Basic Proportionality Theorem

The basic proportionality theorem can be applied to prove part of the mid-point theorem and its converse as discussed below:

A) From the Basic proportionality Theorem or Thales Theorem, we learnt that if a line is drawn parallel to one side of a triangle and intersect the other two sides, then two sides are divided in the same ratio.

This leads us to observe that when the line is drawn from the mid-point of one side and parallel to another side, it bisects the third side.

Proof:
Let triangle \(ABC,\) in which \(D\) is the mid-point of side \(AB.\)
So, \(AD=DB…………(1)\)

Now, if a line \(DE\) is drawn parallel to side \(BC,\) then according to the basic proportionality theorem, can write, 

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

\( \Rightarrow \frac{{AD}}{{DB}} = \frac{{AE}}{{EC}}\)

\( \Rightarrow 1 = \frac{{AE}}{{EC}}\)

\( \Rightarrow AE = EC\)

Thus, \(E\) is the mid-point of side \(AC\)

Hence, the line \(DE\) bisects the side \(AC.\) (proved)

B) Another special case follows from the Converse of Basic Proportionality Theorem which is used to prove the converse of the mid-point partially.

It can be proved that the line joining the mid-points of any two sides of a triangle is parallel to the third side.

Proof:
Let triangle \(ABC,\) in which \(D\) is the mid-point of side \(AB.\)
AD=DB 

So, \(\frac{{AD}}{{DB}} = \frac{{AD}}{{AD}} = 1……..\left( 1 \right)\)

\(E\) is the mid-point of \(AC.\)
\(AE=EC\)

So, \(\frac{{AE}}{{EC}} = \frac{{AE}}{{AE}} = 1……..\left( 2 \right)\)

From \((1,) (2,)\)
ADDB=AEEC

Hence, according to the converse of basic proportionality theorem, we can say that the line \(DE\) is parallel to \(BC.\)

Therefore, \(DE\parallel BC.\) (proved)

Solved Examples – Basic Proportionality Theorem

Q.1. In the triangle \(ABC,DE\parallel BC,\) then find the value of \(EC\)?

Ans:
Given, \(DE\parallel BC.\)
According to the basic proportionality theorem, we know that a line drawn parallel to one side divides the other sides in an equal ratio.
So, \(\frac{{AD}}{{DB}} = \frac{{AE}}{{EC}}\)
\( \Rightarrow \frac{{1.5}}{3} = \frac{1}{{EC}}\)
\( \Rightarrow \frac{1}{2} = \frac{1}{{EC}}\)
\( \Rightarrow EC = {\rm{2}}\)
Hence, the value of \(EC\) is \({\rm{2}}\,{\rm{cm}}{\rm{.}}\)

Q.2. Given, \(DE\parallel AC\) and \(DF\parallel AE,\) prove that \(\frac{{BF}}{{FE}} = \frac{{BE}}{{EC}}\)

Ans:
In triangle \(ABC,DE\parallel AC\)
According to the basic proportionality theorem, we know that a line is drawn parallel to one side divides the remaining sides in an equal ratio.
\(\frac{{BD}}{{DA}} = \frac{{BE}}{{EC}}……….\left( 1 \right)\)
In triangle \(ABE,DF\parallel AE\)
According to the basic proportionality theorem, we know that a line drawn parallel to one side divides the remaining sides in an equal ratio.
\(\frac{{BD}}{{DA}} = \frac{{BF}}{{FE}}……….\left( 2 \right)\)
From \((1)\) and \((2),\)
\(\frac{{BF}}{{FE}} = \frac{{BE}}{{EC}}\)

Q.3. In the triangle \(ABC,\) the values of \(AB = 10\,{\rm{cm}},AC = 12\,{\rm{cm}},\,AD = {\rm{5}}\,{\rm{cm}}\) and \(AE = 6\,{\rm{cm}},\), then check for \(DE\parallel BC.\)

Ans:
Given: of \(AB = 10\,{\rm{cm}},AC = 12\,{\rm{cm}},\,AD = {\rm{5}}\,{\rm{cm}}\) and \(AE = 6\,{\rm{cm}}{\rm{.}}\)
So, \(DB = AB – AD = 10 – 5 = 5\;{\rm{cm}}\)
\(EC = AC – AE = 12 – 6 = 6\;{\rm{cm}}\)
So, \(\frac{{AD}}{{DB}} = \frac{5}{5} = 1…….\left( 1 \right)\)
And, \(\frac{{AE}}{{EC}} = \frac{6}{6} = 1…….\left( 2 \right)\)
From \({\rm{ (1), (2), }}\frac{{AD}}{{DB}} = \frac{{AE}}{{EC}} = 1\)
According to the statement of the converse of basic proportionality theorem, \({\rm{DE}}\parallel {\rm{BC}}{\rm{.}}\)

Q.4. In the triangle \(ABC,DE\parallel BC{\rm{.}}\) then find the value of \(AD\)?

Ans:
Given, \(DE\parallel BC{\rm{.}}\)
According to the basic proportionality theorem, we know that a line drawn parallel to one side divides the other sides in an equal ratio.
So, \(\frac{{AD}}{{DB}} = \frac{{AE}}{{EC}}\)
\( \Rightarrow \frac{{AD}}{{7.2}} = \frac{{1.8}}{{5.4}}\)
\( \Rightarrow \frac{{AD}}{{7.2}} = \frac{1}{3}\)
\( \Rightarrow AD = \frac{{7.2}}{3}\)
\( \Rightarrow AD = 2.4\;{\rm{cm}}\)
Hence, the measure of AD is \(2.4\;{\rm{cm}}{\rm{.}}\)

Q.5. In the given figure \(LM\parallel BC\) and \(LN\parallel CD,\) then prove that \(\frac{{AM}}{{MB}} = \frac{{AN}}{{ND}}.\)

Ans:
In the triangle \(ABC,LM\parallel BC.\)
According to the basic proportionality theorem, we know that a line drawn parallel to one side divides the other sides in an equal ratio.
\(\frac{{AM}}{{MB}} = \frac{{AL}}{{LC}}….\left( 1 \right)\)
In triangle \(ACD,LN\parallel CD.\)
According to the basic proportionality theorem, we know that a line drawn parallel to one side divides the other sides in an equal ratio.
\(\frac{{AN}}{{ND}} = \frac{{AL}}{{LC}}…………\left( 2 \right)\)
From \(\left( 1 \right),\left( 2 \right)\)
\(\frac{{AM}}{{MB}} = \frac{{AN}}{{ND}}\)
Hence, proved.

Summary

In this article, we have learned about the basic proportionality theorem, also known as the Thales theorem, and it is abbreviated as \({\rm{B}}{\rm{.P}}{\rm{.T}}\)

Here, we also discussed the converse of the basic proportionality theorem. Applications of Basic proportionality theorem. Such as how it can be applied to prove mid-point theorem in the triangle and its converse. This will helps us to find the lengths of the sides of the triangles.

Frequently Asked Questions

We have provided some frequently asked questions about basic proportionality theorem here:

Q.1. Define Basic proportionality theorem.
Ans: The Basic proportionality Theorem or Thales Theorem states that when a line is drawn parallel to one side of a triangle and intersect the other two sides, then the line divides the two sides in the same ratio.

Q.2. What is the formula for the Basic proportionality Theorem?
Ans: In a triangle \(ABC,\) such that line \(DE\) has drawn parallel to the base of the triangle \(BC.\) Then, the line \(DE\) divides the sides \(AB, AC\) in the same ratio.
\(\frac{{AD}}{{DB}} = \frac{{AE}}{{EC}}\)

Q.3. What does \({\rm{B}}{\rm{.P}}{\rm{.T}}\) mean?
Ans: The Basic proportionality theorem abbreviated as \({\rm{B}}{\rm{.P}}{\rm{.T}}{\rm{.}}\) It is also called the Thales theorem.

Q.4. What are the properties of the basic proportionality theorem?
Ans: The Basic proportionality Theorem is a theorem dealing with ratios of intercepts made by a line on two of its sides under different circumstances. It has no property as such, but this theorem has a lot of applications in geometry.

Q.5. What is the application of the basic proportionality theorem?
Ans: The basic proportionality theorem helps to find the lengths of two sides. One of the applications of the basic proportionality theorem is finding the relationship of sides of two equiangular triangles.

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