: 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. 5.8(i)].

Fig. 5.8

Here, you need to do some construction. Using Euclid’s Postulate 3, you can draw a

circle with point A as the centre and AB as the radius [see Fig. 5.8(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. 5.8 (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.

How Can AB and AC are the radii of the same circle And AB And BC be the radii of the same circle?

If you want diagrams refer page no.84 in NCERT textbook

Consider the circle drawn by taking A as centre and radius AB. It is clear that point C lies on it (as C is the point of the intersection of two circles). Join AC. Thus, AB and AC are the two radii of the same circle. Hence, they are equal.

Now, consider the circle drawn by taking B as centre and radius AB. It is clear that point C lies on it as well. Join BC. Thus, BC and AC are the two radii of the same circle. Hence, they are equal.

Hope now the question will be cleared to you.

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