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Prove that an equilateral triangle can be constracted on any given line segment

Answer»

Solution :Let AB be the line segment of give LENGTH with A as CENTRE and AB as the radius DRAW a circle with B as centre and BA as the radius draw another circle CUTTING the first circle at C . Join AC and BC to form `triangle ABC`

Now AB=AC(radii of the same circle)
And BA =BC(radii of the same circle)
`therefore` AB=BC`[therefore bar(BA)=bar(AB)]`
But by Euclid 's Axiom 1 it follows that the things which are equal to the same things are equal to one another `therefore` AB=BC=AC
Hence `triangle` ABC is an equilateral triangle


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