Example (Revised) - Chapter 5 - Introduction To Euclids Geometry - Ncert Solutions class 9 - Maths

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NCERT Solutions for Class 9 Maths Chapter 5 - Introduction to Euclid's Geometry

Example 1:

If A, B and C are three points on a line, and B lies between A and C (see Fig.), then prove that $\mathrm{AB}+\mathrm{BC}=\mathrm{AC}$.

Solution :

In the figure given above, $\mathrm{AC}$ coincides with $\mathrm{AB}+\mathrm{BC}$.
Also, Euclid's Axiom (4) says that things which coincide with one another are equal to one another. So, it can be deduced that
$
\mathrm{AB}+\mathrm{BC}=\mathrm{AC}
$

Note that in this solution, it has been assumed that there is a unique line passing through two points.
Example 2 :

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

In the statement above, a line segment of any length is given, say $A B$ [see Fig.].

Here, you need to do some construction. Using Euclid's Postulate 3, you can draw a circle with point $\mathrm{A}$ as the centre and $\mathrm{AB}$ as the radius [see Fig. 5.8(ii)]. Similarly, draw another circle with point $\mathrm{B}$ as the centre and $\mathrm{BA}$ as the radius. The two circles meet at a point, say C. Now, draw the line segments $\mathrm{AC}$ and $\mathrm{BC}$ to form $\triangle \mathrm{ABC}$ [see Fig. 5.8 (iii)].
So, you have to prove that this triangle is equilateral, i.e., $\mathrm{AB}=\mathrm{AC}=\mathrm{BC}$.
Now, $\quad \mathrm{AB}=\mathrm{AC}$, since they are the radii of the same circle
Similarly, $\mathrm{AB}=\mathrm{BC} \quad$ (Radii of the same circle)
From these two facts, and Euclid's axiom that things which are equal to the same thing are equal to one another, you can conclude that $\mathrm{AB}=\mathrm{BC}=\mathrm{AC}$.
So, $\triangle \mathrm{ABC}$ is an equilateral triangle.
Note that here Euclid has assumed, without mentioning anywhere, that the two circles drawn with centres $\mathrm{A}$ and $\mathrm{B}$ will meet each other at a point.