Basic Proportionality Theorem (Thales Theorem) & its Converse

The Basic Proportionality Theorem (BPT), also known as Thales’ Theorem, is a fundamental concept in geometry, specifically dealing with triangles. It establishes a relationship between the sides of a triangle when a line is drawn parallel to one side and intersecting the other two sides.

The converse of the BPT works in reverse. If a line divides any two sides of a triangle in the same ratio, then the line is parallel to the third side.

Formulae

Basic Proportionality Theorem (BPT):

If a line is drawn parallel to one side of a triangle intersecting the other two sides in distinct points, then the other two sides are divided in the same ratio.

In triangle $ABC$, if $DE \parallel BC$, then:

$\frac{AD}{DB} = \frac{AE}{EC}$

Converse of BPT:

If a line divides any two sides of a triangle in the same ratio, then the line is parallel to the third side.

In triangle $ABC$, if $\frac{AD}{DB} = \frac{AE}{EC}$, then $DE \parallel BC$.

Examples

Example-1:

In triangle $PQR$, $ST$ is drawn parallel to $QR$. Given that $PS = 3 \text{ cm}$, $SQ = 6 \text{ cm}$, and $PT = 4 \text{ cm}$. Find $TR$.

Solution:

By BPT, $\frac{PS}{SQ} = \frac{PT}{TR}$

Substituting the given values: $\frac{3}{6} = \frac{4}{TR}$

Cross-multiplying: $3 \times TR = 6 \times 4$

$TR = \frac{24}{3} = 8 \text{ cm}$

Example-2:

In triangle $XYZ$, $AB$ is drawn such that $XA = 4 \text{ cm}$, $AY = 5 \text{ cm}$, $XB = 8 \text{ cm}$, and $BZ = 10 \text{ cm}$. Is $AB \parallel YZ$? Justify your answer.

Solution:

We check if the converse of BPT holds. $\frac{XA}{AY} = \frac{4}{5}$ and $\frac{XB}{BZ} = \frac{8}{10} = \frac{4}{5}$

Since $\frac{XA}{AY} = \frac{XB}{BZ}$, the converse of BPT holds. Therefore, $AB \parallel YZ$.

Theorem with Proof

Theorem: Basic Proportionality Theorem (BPT) – If a line is drawn parallel to one side of a triangle intersecting the other two sides in distinct points, then the other two sides are divided in the same ratio.

Proof:

Let’s consider a triangle $ABC$. Let $DE$ be a line drawn parallel to $BC$ and intersecting $AB$ at $D$ and $AC$ at $E$. We need to prove that $\frac{AD}{DB} = \frac{AE}{EC}$.

1. **Construction:** Join $BE$ and $CD$. Draw $EM \perp AB$ and $DN \perp AC$.

2. **Proof:**

Area of triangle $ADE$, $A(ADE) = \frac{1}{2} \times AD \times EM$ (Using $AD$ as the base)

Also, $A(ADE) = \frac{1}{2} \times AE \times DN$ (Using $AE$ as the base)

Area of triangle $BDE$, $A(BDE) = \frac{1}{2} \times DB \times EM$

Area of triangle $CED$, $A(CED) = \frac{1}{2} \times EC \times DN$

3. **Ratio Calculation:**

$\frac{A(ADE)}{A(BDE)} = \frac{\frac{1}{2} \times AD \times EM}{\frac{1}{2} \times DB \times EM} = \frac{AD}{DB}$ (1)

$\frac{A(ADE)}{A(CED)} = \frac{\frac{1}{2} \times AE \times DN}{\frac{1}{2} \times EC \times DN} = \frac{AE}{EC}$ (2)

4. **Equal Areas:** Triangles $BDE$ and $CED$ are on the same base $DE$ and between the same parallels $DE$ and $BC$. Therefore, their areas are equal, $A(BDE) = A(CED)$.

5. **Conclusion:**

From (1) and (2), we have: $\frac{A(ADE)}{A(BDE)} = \frac{A(ADE)}{A(CED)}$

Since the denominators are equal ($A(BDE) = A(CED)$), we get $\frac{AD}{DB} = \frac{AE}{EC}$.

Hence, the Basic Proportionality Theorem (BPT) is proved.

Common mistakes by students

• Incorrectly applying the ratios: Students sometimes mix up the corresponding sides in the ratios. For example, they might write $\frac{AD}{AB} = \frac{AE}{AC}$ instead of $\frac{AD}{DB} = \frac{AE}{EC}$ or $\frac{AD}{AB} = \frac{AE}{AC}$.
• Not recognizing the parallel lines: Students may fail to identify the line segments as parallel, leading to the incorrect application of the theorem.
• Misinterpreting the converse: They might assume the converse holds even when the ratios are not equal. It’s crucial to verify the ratios before concluding that the line is parallel.
• Forgetting to consider the complete sides of the triangle when ratio is needed.

Real Life Application

The Basic Proportionality Theorem is used in various real-life applications, including:

• Architecture and Construction: For designing structures and ensuring that parallel lines are maintained in buildings, bridges, and other constructions.
• Mapping and Surveying: Used for calculating distances and heights, particularly in situations where direct measurement is difficult.
• Photography and Art: The theorem helps in understanding perspective and the proportions of objects in a scene.
• Computer Graphics: Used for rendering 3D objects onto 2D screens by scaling and positioning them correctly.

Fun Fact

Thales of Miletus, the Greek mathematician who is often credited with the discovery of this theorem, is said to have used it to measure the height of the pyramids in Egypt! He used similar triangles formed by the pyramid’s shadow and his own shadow to calculate its height.

Recommended YouTube Videos for Deeper Understanding

Q.1 If in triangle $ABC$, $DE \parallel BC$, $AD = 3$ cm, $DB = 4$ cm, and $AE = 6$ cm, then find the length of $EC$.

Check Solution

Ans: B

Using Basic Proportionality Theorem, $\frac{AD}{DB} = \frac{AE}{EC}$. Substituting the given values, $\frac{3}{4} = \frac{6}{EC}$. Therefore, $EC = \frac{6 \times 4}{3} = 8$ cm.

Q.2 In triangle $PQR$, $ST$ is drawn such that $ST \parallel QR$. If $PS = x$, $SQ = x-2$, $PT = x+2$, and $TR = x-1$, then find the value of $x$.

Check Solution

Ans: B

Using Basic Proportionality Theorem, $\frac{PS}{SQ} = \frac{PT}{TR}$. Substituting the given values, $\frac{x}{x-2} = \frac{x+2}{x-1}$. Cross-multiplying, $x(x-1) = (x+2)(x-2)$. This simplifies to $x^2 – x = x^2 – 4$. Therefore, $x = 4$.

Q.3 In triangle $LMN$, $XY$ is a line segment parallel to $MN$. If $LX = 2$, $XM = 3$, and $LN = 6$, then find the length of $LY$.

Check Solution

Ans: A

Using Basic Proportionality Theorem, $\frac{LX}{XM} = \frac{LY}{YN}$. Also, $\frac{LX}{LM} = \frac{LY}{LN}$. Thus, $\frac{2}{2+3} = \frac{LY}{6}$. Solving for $LY$, $LY = \frac{2 \times 6}{5} = 2.4$.

Q.4 In $\triangle ABC$, $DE$ is drawn such that $AD = 4$ cm, $DB = 4.5$ cm, $AE = 8$ cm, and $EC = 9$ cm. Is $DE \parallel BC$?

Check Solution

Ans: A

We check if the ratios are equal: $\frac{AD}{DB} = \frac{4}{4.5} = \frac{8}{9}$, and $\frac{AE}{EC} = \frac{8}{9}$. Since $\frac{AD}{DB} = \frac{AE}{EC}$, by the converse of the Basic Proportionality Theorem, $DE \parallel BC$.

Q.5 In $\triangle XYZ$, line segment $AB$ intersects $XY$ at $A$ and $XZ$ at $B$. Given $XA = 4$, $AY = 8$, $XB = 5$, and $BZ = 10$, then $AB$ is:

Check Solution

Ans: A

We check if the ratios are equal: $\frac{XA}{AY} = \frac{4}{8} = \frac{1}{2}$, and $\frac{XB}{BZ} = \frac{5}{10} = \frac{1}{2}$. Since $\frac{XA}{AY} = \frac{XB}{BZ}$, by the converse of the Basic Proportionality Theorem, $AB \parallel YZ$.

Next Topic: Areas of Similar Triangles: Ratio Property