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Introduction to Euclid's Geometry: Axioms and Postulates - Trace historical development of geometry and major civilisational contributions

Grade 9CBSE

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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The term 'Geometry' is derived from the Greek words 'geo' (earth) and 'metrein' (to measure). Ancient civilizations used geometry for practical needs like land measurement, architecture, and irrigation. In Egypt, geometry was essential for restoring boundaries after the Nile's annual flooding, while in the Indus Valley Civilization (Harappa and Mohenjo-Daro), bricks were produced in a fixed ratio of 4:2:14:2:1 for length, breadth, and thickness.

A rectangular brick representing the standardized dimensions used in the Indus Valley Civilisation.
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Euclid, a teacher of mathematics at Alexandria, organized the known geometrical knowledge of his time into a treatise called 'The Elements'. He divided it into thirteen chapters, each called a 'book'. He began his work by defining basic terms such as point, line, and surface, which he used to build his system of axioms and postulates.

Flowchart showing the organization of Euclid's Elements.
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Euclid's Postulate 1 states that a straight line may be drawn from any one point to any other point. This is supplemented by the axiom that through two distinct points, there is a unique line passing through them.

A unique straight line passing through two distinct points P and Q.
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Euclid's Postulate 5 (Parallel Postulate) states that if a straight line falling on two straight lines makes the interior angles on the same side less than two right angles (180∘180^\circ), the two straight lines, if produced indefinitely, meet on that side where the sum of angles is less than two right angles.

Illustration of Euclid's fifth postulate showing two lines converging on the side where the sum of interior angles is less than 180 degrees.

📐Formulae

a=b,c=d  ⟹  a+c=b+da = b, c = d \implies a + c = b + d

a=b,c=d  ⟹  a−c=b−da = b, c = d \implies a - c = b - d

If a=b, then 2a=2b\text{If } a = b, \text{ then } 2a = 2b

If a=b, then a2=b2\text{If } a = b, \text{ then } \frac{a}{2} = \frac{b}{2}

∠1+∠2<180∘  ⟹  Lines will intersect on that side.\angle 1 + \angle 2 < 180^\circ \implies \text{Lines will intersect on that side.}

💡Examples

Problem 1:

If a point CC lies between two points AA and BB such that AC=BCAC = BC, then prove that AC=12ABAC = \frac{1}{2} AB.

Solution:

Given AC=BCAC = BC. Adding ACAC to both sides (Euclid's Axiom 2: If equals are added to equals, the wholes are equal): AC+AC=BC+ACAC + AC = BC + AC 2AC=BC+AC2AC = BC + AC Since BC+ACBC + AC coincides with ABAB (Euclid's Axiom 4: Things which coincide with one another are equal to one another): 2AC=AB2AC = AB Dividing both sides by 22 (Euclid's Axiom 7: Things which are halves of the same things are equal to one another): AC=12ABAC = \frac{1}{2} AB

Explanation:

This proof utilizes Euclid's Axioms regarding addition and halves to show the relationship between the segments of a bisected line.

Problem 2:

Given two distinct points AA and BB, is there at least one line that passes through them?

Solution:

Yes, according to Euclid's Postulate 1: 'A straight line may be drawn from any one point to any other point.' Modern geometry refines this as an axiom stating: 'Given two distinct points, there is a unique line that passes through them.'

Explanation:

This is a direct application of Euclid's first postulate, asserting the existence of a path between two points.

Problem 3:

Consider the following statement: There exists a pair of straight lines that are everywhere equidistant from one another. Is this statement a direct consequence of Euclid’s fifth postulate? Explain.

Solution:

Yes. This statement is equivalent to the existence of parallel lines. If the distance between two lines is constant, they will never meet, which corresponds to the case where the sum of interior angles on the same side is exactly 180∘180^\circ (22 right angles).

Explanation:

This is a reinterpretation of the Parallel Postulate (Postulate 5) through the lens of equidistance.

Problem 4:

In the figure, if AC=BDAC = BD, then prove that AB=CDAB = CD.

A line segment AD with points B and C marked in between in order.

Solution:

  1. We are given AC=BDAC = BD.
  2. From the figure, ACAC can be written as AB+BCAB + BC and BDBD can be written as BC+CDBC + CD.
  3. Substituting these in the given equation: AB+BC=BC+CDAB + BC = BC + CD.
  4. According to Euclid's axiom, 'If equals are subtracted from equals, the remainders are equal'. Subtracting BCBC from both sides: AB+BC−BC=BC+CD−BCAB + BC - BC = BC + CD - BC AB=CDAB = CD

Explanation:

This problem uses Euclid's axiom about subtraction of equals to show the relationship between overlapping line segments on a straight line.

Problem 5:

Show that an equilateral triangle can be constructed on any given line segment.

Construction of an equilateral triangle using two intersecting circles with centers at the endpoints of a line segment.

Solution:

  1. Let ABAB be the given line segment.
  2. Using Euclid's Postulate 3, draw a circle with center AA and radius ABAB.
  3. Draw another circle with center BB and radius BABA.
  4. Let the two circles intersect at point CC.
  5. Draw line segments ACAC and BCBC (Postulate 1).
  6. Now, AB=ACAB = AC (radii of the same circle) and AB=BCAB = BC (radii of the same circle).
  7. By Euclid's Axiom 1 (Things which are equal to the same thing are equal to one another), AB=AC=BCAB = AC = BC.
  8. Therefore, △ABC\triangle ABC is an equilateral triangle.

Explanation:

This construction demonstrates the application of Euclid's postulates regarding circles and lines, combined with his first axiom regarding equality.