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 AB + BC = AC. euclid

Solution :-
 In the figure given above, AC coincides with AB + 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 AB + BC = 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 AB [see Fig.(ii)]
 Construction :- Using Euclid’s Postulate 3,  draw a circle with point A as the centre and AB as the radius [see Fig.(ii)]. 
 Similarly, draw another circle with point B as the centre and BA as the radius. The two circles meet at a point, say C. 
 Now, draw the line segments AC and BC to form Δ ABC [see Fig. (iii)]. 
 So, you have to prove that this triangle is equilateral, i.e., AB = AC = BC. 
 Now, AB = AC, since they are the radii of the same circle (1) Similarly, AB = BC (Radii of the same circle) (2) 
 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 AB = BC = AC. 
 So, Δ ABC is an equilateral triangle.

Example-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 :-
 Take any line l and a point P not on l. 
 Then, by Playfair’s axiom, which is equivalent to the fifth postulate, 
 we know that there is a unique line m through P which is parallel to l. 
 Now, the distance of a point from a line is the length of the perpendicular from the point to the line.
 This distance will be the same for any point on m from l and any point on l from m. 
 So, these two lines are everywhere equidistant from one another. 

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