Question
Prove that an equilateral triangle can be constructed on any given line segment.
✓
Answer
In the statement above, a line segment of any length is given, say $AB [$see Fig. $(i)].$
Here, we need to do some construction. Using Euclid’s Postulate 3, we can 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 $C$. Then, draw the line segments $AC$ and $BC$ to form $\triangle ABC [$see Fig. $(iii)].$
Therefore, we have to prove that this triangle is equilateral, i.e., $AB = AC = BC.$
Now, $AB = AC,$ because 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, we can conclude that $AB = BC = AC.$ Therefore, $\triangle$ ABC is an equilateral triangle.
Hence,proved.
Here, we need to do some construction. Using Euclid’s Postulate 3, we can 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 $C$. Then, draw the line segments $AC$ and $BC$ to form $\triangle ABC [$see Fig. $(iii)].$
Therefore, we have to prove that this triangle is equilateral, i.e., $AB = AC = BC.$
Now, $AB = AC,$ because 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, we can conclude that $AB = BC = AC.$ Therefore, $\triangle$ ABC is an equilateral triangle.
Hence,proved.
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