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Introduction to Euclid's Geometry - Euclid's Definitions, Axioms, and Postulates

Grade 9CBSE

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

🔑Concepts

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Euclid's Definitions: Euclid defined basic geometric terms to build his system. A point is that which has no part, visually represented as a position with no dimensions. A line is breadthless length, appearing as a straight path extending in both directions. A surface is that which has length and breadth only, like a flat plane without thickness.

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Axioms vs. Postulates: Euclid distinguished between assumptions. Axioms (or common notions) are basic assumptions used throughout mathematics and not just in geometry, such as 'Things which are equal to the same thing are also equal to one another'. Postulates are assumptions specific to the field of geometry, such as the ability to draw a straight line between two points.

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Euclid's Five Postulates: These form the foundation of Euclidean geometry. 1. A straight line may be drawn from any one point to any other point. 2. A terminated line (line segment) can be produced indefinitely. 3. A circle can be drawn with any centre and any radius. 4. All right angles are equal to one another. 5. The Parallel Postulate, which dictates when two lines will eventually intersect.

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The Fifth Postulate (Parallel Postulate): If a straight line falling on two straight lines makes the interior angles on the same side taken together less than two right angles (180∘180^\circ), then the two straight lines, if produced indefinitely, meet on that side on which the sum of angles is less than two right angles. Visually, imagine two lines ll and mm intersected by a transversal nn; if the sum of interior angles ∠1+∠2<180∘\angle 1 + \angle 2 < 180^\circ, the lines converge and cross on that side.

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Playfair's Axiom: This is an equivalent version of Euclid's fifth postulate. It states that for every line ll and for every point PP not lying on ll, there exists a unique line mm passing through PP and parallel to ll. Visually, this means only one distinct line can be drawn through a specific point that will never meet a given line, no matter how far they are extended.

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The Whole and the Part: One of Euclid's axioms states 'The whole is greater than the part'. Visually, if a line segment ABAB contains a point CC such that AC+CB=ABAC + CB = AB, then the length of ABAB is always greater than the length of ACAC or CBCB. This logic applies to areas and volumes as well.

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Consistency and Non-Euclidean Geometry: Euclid's first four postulates are simple and intuitive, but the fifth is complex. For centuries, mathematicians tried to prove the fifth postulate using the first four. This failure led to the discovery of Non-Euclidean geometries, where the fifth postulate does not hold (e.g., spherical geometry where parallel lines do not exist).

📐Formulae

If a=ba = b and c=dc = d, then a+c=b+da + c = b + d (Adding equals to equals)

If a=ba = b and c=dc = d, then a−c=b−da - c = b - d (Subtracting equals from equals)

If a=ba = b and b=cb = c, then a=ca = c (Transitive property of equality)

If x=yx = y, then 2x=2y2x = 2y (Doubles of equals are equal)

If x=yx = y, then x2=y2\frac{x}{2} = \frac{y}{2} (Halves of equals are equal)

Sum of interior angles <180∘  ⟹  < 180^\circ \implies Lines intersect

💡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. Explain by drawing the figure.

Solution:

  1. We are given AC=BCAC = BC.
  2. According to Euclid's axiom, 'if equals are added to equals, the wholes are equal'.
  3. Let us add ACAC to both sides of the given equation: AC+AC=BC+ACAC + AC = BC + AC.
  4. This gives 2AC=BC+AC2AC = BC + AC.
  5. From the visual representation of the line segment, the point CC lies between AA and BB, so AC+BCAC + BC coincides with the line segment ABAB.
  6. According to the axiom 'things which coincide with one another are equal to one another', we have AC+BC=ABAC + BC = AB.
  7. Therefore, 2AC=AB2AC = AB.
  8. Dividing both sides by 22 (or taking halves), we get AC=12ABAC = \frac{1}{2} AB.

Explanation:

This solution utilizes Euclid's Axiom 2 (adding equals to equals) and Axiom 4 (coinciding objects) to establish a relationship between a part of a segment and the whole segment.

Problem 2:

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

Solution:

  1. Let ABAB be the given line segment.
  2. Using Euclid's Postulate 3 (a circle can be drawn with any centre and radius), draw a circle with centre AA and radius ABAB.
  3. Similarly, draw another circle with centre BB and radius BABA.
  4. Let these two circles intersect at a point CC.
  5. Draw line segments ACAC and BCBC (Postulate 1).
  6. Now, AB=ACAB = AC (radii of the same circle).
  7. Also, AB=BCAB = BC (radii of the same circle).
  8. According to Euclid's Axiom 1 (things equal to the same thing are equal to each other), since AC=ABAC = AB and BC=ABBC = AB, then AC=BCAC = BC.
  9. Thus, AB=BC=ACAB = BC = AC. Since all three sides are equal, △ABC\triangle ABC is an equilateral triangle.

Explanation:

This proof demonstrates the application of Euclid's Postulates 1 and 3 along with Axiom 1 to construct a geometric figure based on fundamental assumptions.