Basic Proportionality Theorem and Equal Intercept Theorem

Basic Proportionality Theorem: If a line is parallel to a side of a triangle which intersects the other sides into two distinct points, then the line divides those sides in proportion. Let’s not stop at the statement, we need to find a proof that its true. So shall we begin?

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Introduction

Basic Proportionality Theorem was first stated by Thales, a Greek mathematician. Hence it is also known as Thales Theorem. Thales first initiated and formulated the Theoretical Study of Geometry to make astronomy a more exact science. What is this theorem that Thales found important for his study of astronomy? Let us find it out.

Basic Proportionality Theorem (can be abbreviated as BPT) states that, if a line is parallel to a side of a triangle which intersects the other sides into two distinct points, then the line divides those sides in proportion. 

 

In the figure alongside, if we consider DE is parallel to BC, then according to the theorem,

\(\frac{AD}{BD} = \frac{AE}{CE} \)

Let’s not stop at the statement, we need to find a proof that its true. So shall we begin?

Browse more Topics under Triangles

• Properties of Triangles
• Congruent Triangles
• Similarity of Triangles
• Inequalities of Triangles
• Pythagoras Theorem and its Applications

Download Triangles Cheat Sheet PDF

PROOF OF BPT

Given: In  ΔABC, DE is parallel to BC

Line DE intersects sides AB and AC in points D and E respectively.

To Prove: \( \frac{AD}{BD} = \frac{AE}{CE} \)

Construction: Draw EF ⟂ AD and DG⟂ AE and join the segments BE and CD.

Proof: 

Area of Triangle= ½ × base × height

In ΔADE and ΔBDE,

$$\frac{Ar(ADE)}{Ar(DBE)} =\frac{\frac{1}{2}\times AD \times EF}{\frac{1}{2}\times DB \times EF} = \frac{AD}{DB} \hspace{1.4cm} (1)$$

In ΔADE and ΔCDE,

$$\frac{Ar(ADE)}{Ar(ECD)} =\frac{\frac{1}{2}\times AE \times DG}{\frac{1}{2}\times EC \times DG} = \frac{AE}{EC} \hspace{1.4cm} (2)$$

Note that ΔDBE and ΔECD have a common base DE and lie between the same parallels DE and BC. Also, we know that triangles having the same base and lying between the same parallels are equal in area.

So, we can say that
$$Ar(\Delta DBE) = Ar(\Delta ECD)$$

Therefore,

\( \frac{A(ΔADE)}{A(ΔBDE)} = \frac{A(ΔADE)}{A(ΔCDE)} \)

Therefore,

\(\frac{AD}{BD} = \frac{AE}{CE} \)

Hence Proved.

The BPT also has a converse which states, if a line divides any two sides of a triangle in the same ratio, then the line is parallel to the third side.

(Note: A converse of any theorem is just a reverse of the original theorem, just like we have active and passive voices in English.)

Read the properties of Triangles and Quadrilaterals here.

PROPERTIES OF BPT

The BPT has 2 properties.

1. Property of an angle bisector.
2. Property of Intercepts made by three parallel lines on a transversal.

Property of an Angle Bisector

Statement: In a triangle, the angle bisector divides the side opposite to the angle in the ratio of the remaining sides.

In the given figure, seg AD is the angle bisector of ∠BAC.

According to the property,

\( \frac{BD}{DC}=\frac{AB}{AC} \)

Property of Intercepts made by three parallel lines on a transversal

Statement: The ratio of the intercepts made on the transversal by three parallel lines is equal to the ratio of the corresponding intercepts made on any other transversal of the same parallel line.

Consider the above figure,  line l, m, and n are parallel to each other. Transversals p and q intersect the lines at point A, B, C and D, E, F. So according to the property,

\( \frac{AB}{BC}=\frac{DE}{EF} \)

Learn more about Similarity of Triangles here.

Solved Example

Question 1: In triangle ABC, seg AD is the angle bisector of∠BAC. BD=6, DC=8, AB=15 Find AC.

Answer :

segments AD bisects ∠BAC     (given)

.’. AB/AC=BD/DC                      (Angle bisector property)

Assume AC= x

.’. 15/x=6/8

.’. 15 × 8 = 6 × x

.’.  x= (15 × 8) / 6 = 20.

Therefore, AC = 20.

Question 2: We are given that in triangle PQR, MN intersects PQ and PR at M and N respectively such that PM = 3 cm, MQ = 9 cm, PN = 2 cm and NR = 6 cm. Is MN parallel to QR?

Answer :