In geometry, triangles are similar when they share the same shape, irrespective of their size. This happens when:

  1. Corresponding angles are equal.
  2. Corresponding sides are in the same ratio (proportion).

For example, if \boldsymbol, then:

\boldsymbol \displaystyle\boldsymbol

This concept is foundational in geometry and has practical applications in fields such as construction, navigation, and design.

Basic Proportionality Theorem (Thales' Theorem)

Statement

If a line is drawn parallel to one side of a triangle, intersecting the other two sides, the line divides those sides in the same ratio.

In \boldsymbol, if \boldsymbol, then:

\displaystyle\boldsymbol

Understanding Thales' Theorem

Setup

  1. Draw any angle \boldsymbol and on one arm \boldsymbol, mark points \boldsymbol such that: \boldsymbol
  2. Draw another arm \boldsymbol and mark a point \boldsymbol on it.
  3. From \boldsymbol, draw \boldsymbol.

Observations

Measure \boldsymbol, \boldsymbol, \boldsymbol, and \boldsymbol.
You will find:

\displaystyle\boldsymbol{\frac = \frac}

This supports the Basic Proportionality Theorem.

Proof of Basic Proportionality Theorem

Given

In \boldsymbol, \boldsymbol intersects \boldsymbol and \boldsymbol at \boldsymbol and \boldsymbol.

To Prove

\displaystyle\boldsymbol{\frac = \frac}

Construction

  1. Draw \boldsymbol and \boldsymbol.

Proof

  1. Area of \boldsymbol is proportional to its base \boldsymbol:
    \displaystyle\boldsymbol
  2. Similarly, for \boldsymbol:
    \displaystyle\boldsymbol
  3. Taking the ratio of areas:
    Area of \displaystyle\boldsymbol{\frac{\text{Area of } \triangle ADE}{\text{Area of } \triangle BDE} = \frac}
  4. Similarly, for \boldsymbol and \boldsymbol:
    Area of \displaystyle\boldsymbol{\frac{\text{Area of } \triangle ADE}{\text{Area of } \triangle DEC} = \frac}
  5. Thus, \displaystyle\boldsymbol{\frac = \frac}.

Converse of Basic Proportionality Theorem

Statement

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

Proof

  1. Suppose \displaystyle\boldsymbol{\frac = \frac}.
  2. Assume \boldsymbol is not parallel to \boldsymbol.
  3. Draw \boldsymbol \parallel BC}, and by the Basic Proportionality Theorem: \displaystyle\boldsymbol{\frac = \frac{AE'}{E'C}}
  4. But this contradicts \displaystyle\boldsymbol{\frac = \frac}. Hence, \boldsymbol.

Examples Based on Theorems

Example 1: Prove \displaystyle\boldsymbol{\frac = \frac = \frac}

Solution:
From Thales' Theorem:

\displaystyle\boldsymbol

Adding 1 to both sides:

\displaystyle\boldsymbol \displaystyle\boldsymbol

Thus:

\displaystyle\boldsymbol

Example 2: Prove \displaystyle\boldsymbol in a trapezium.

Solution:
Join \boldsymbol, intersecting \boldsymbol at \boldsymbol.

  1. Using Thales' Theorem in \boldsymbol: \displaystyle\boldsymbol
  2. Similarly, in \boldsymbol: \displaystyle\boldsymbol

Thus:

\displaystyle\boldsymbol

Example 3: Prove \boldsymbol.

Solution:

  1. \boldsymbol, so by Thales' Theorem: \displaystyle\boldsymbol
  2. \boldsymbol (corresponding angles).

Thus, \boldsymbol.

Practice Problems

  1. In \boldsymbol, prove that \displaystyle\boldsymbol when \boldsymbol.
  2. Prove that the ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding sides.
  3. In \boldsymbol, \boldsymbol. If \boldsymbol cm, \boldsymbol cm, and \boldsymbol cm, find \boldsymbol.

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