Let line AB be the given finite straight line. It is required to construct an equilateral triangle on the straight line AB.

Describe the circle BCD with center A and radius AB.

Again describe the circle ACE with center B and radius BA.

Join the straight lines line AC and line BC from the point C at which the circles cut one another to the points point A and point B.

Now, since the point A is the center of the circle BCD, therefore line AC equals line AB. Again, since the point B is the center of the circle ACE, therefore line BC equals line AB. But line AC was proved equal to line AB, therefore each of the straight lines line AC and line BC equals line AB. And things which equal the same thing also equal one another, therefore line AC also equals line BC. Therefore the three straight lines line AC, line AB, and line BC equal one another. Therefore the triangle ABC is equilateral, and it has been constructed on the given finite straight line AB.

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