**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.Â **

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:**Â InÂ Î”ABC, DE is parallel to BC

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

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

**Construction:Â **Draw EF âŸ‚ AD and DGâŸ‚ AE and join the segments BE and CD.

**Proof:Â **

Area of Triangle=Â Â½Â Ã— base Ã— height

InÂ Î”ADE andÂ Î”BDE,

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

InÂ Î”ADE andÂ Î”CDE,

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

Note that Î”DBE and Î”ECD 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,

\( \frac{A(Î”ADE)}{A(Î”BDE)} = \frac{A(Î”ADE)}{A(Î”CDE)} \)

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 BP****T**

The BPT has 2 properties.

- Property of an angle bisector.
- 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 ofÂ âˆ 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,Â 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 triangleÂ ABC, seg AD is the angle bisector ofâˆ BAC. BD=6, DC=8, AB=15 Find AC.**

**Answer :**

segments AD bisectsÂ âˆ BACÂ Â Â (given)

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

Assume AC= x

.’. 15/x=6/8

.’. 15Â Ã— 8 = 6 Ã— x

.’.Â x= (15Â Ã— 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 :**

Hence, by the converse of basic proportionality theorem, we have MN parallel to QR.**Question 3: What is BPT in triangles?**

**Answer:** This theorem states that, if you draw a line is parallel to a side of a triangle that transects the other sides into two distinct points then the line divides those sides in proportion. In addition, if a line divides any two sides of a triangle in the same ratio, then the line is parallel to the third side.

**Question 4: What is the corollary of BPT?**

**Answer:** It is a transversal that is parallel to a side in a triangle that defines a new smaller triangle that is similar to the original triangle. In it proof the corresponding angles of the two triangles are congruent.

**Question 5: What is a triangles similarity theorem?**

**Answer:** Triangle similarity theorem identifies that under which conditions triangles are similar. In addition, if two of the angles are the same, then the third angle is the same and the triangles are similar. Furthermore, if two sides are in the same proportion and the included angles are the same then the triangles are similar.

**Question 6: Are parallel triangles similar?**

**Answer:** If the procedures of the equivalent sides of two triangles are proportional then the triangles are similar. Similarly, if the measure of two sides in one triangle is proportional to the same sides and angles of another triangle then the triangles are similar.

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