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Q6:

In Fig. 5.10, if $AC = BD$, then prove that $AB = CD$.

Solution :

Step 1: Identifying Given Parameters and Geometric Context

Given the linear arrangement of points $A, B, C,$ and $D$ on a single line segment as shown in the geometric configuration:

  • $AC = BD$
  • Points $A, B, C, D$ are collinear and ordered such that $B$ lies between $A$ and $C$, and $C$ lies between $B$ and $D$.

[Per Euclid’s Axioms, specifically the segment addition postulate, the length of a segment composed of adjacent parts is the sum of those parts.]

A B C D

Step 2: Formulating the Algebraic Equations

Based on the relative positions of the points, we express segments $AC$ and $BD$ as the sums of their constituent segments:

$AC = AB + BC$ [Equation 1]

$BD = BC + CD$ [Equation 2]

Step 3: Applying Euclid's Axioms

We are given the equality $AC = BD$. By substituting the expressions from Equation 1 and Equation 2 into this equality, we obtain:

$AB + BC = BC + CD$

Step 4: Algebraic Reduction

To isolate $AB$ and $CD$, we apply Euclid’s Third Common Notion: "If equals be subtracted from equals, the remainders are equal."

Subtracting the segment $BC$ from both sides of the equation:

$(AB + BC) - BC = (BC + CD) - BC$

$AB = CD$

Step 5: Theoretical Justification

The derivation holds because $BC$ is a common length present in both original segments $AC$ and $BD$. By removing the overlapping segment, the remaining non-overlapping parts $AB$ and $CD$ must necessarily be of equal magnitude.

Final Solution: $AB = CD$ (Proved)


More Questions from Class 9 Mathematics Introduction to Euclid’s Geometry EXERCISE 5.1


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