construct an equilateral triangle on a segment

I.1

On a given finite straight line to construct an equilateral triangle.

—Euclid

construct an equalateral triangle on a segment

from the given points \(A\) and \(B\)

set segment \(\bar{AB}\).

point \(A\)

point \(B\)

PROBLEM: construct an equilateral triangle on the segment \(\bar{AB}\).

construct circle \((A, B)\) as \(c_1\).

construct circle \((B, A)\) as \(c_2\).

Todo

define intersection point

identify the two intersection points of circles \(c_1\) and \(c_2\) as points \(C\) and \(D\)

from the points \(A\) and \(C\)

set segment \(\bar{AC}\).

from the points \(B\) and \(C\)

set segment \(\bar{BC}\).

As the circle is defined, all points on the perimeter are an equal distance from the center.

Now, since the point \(A\) is the center of the circle \(c_1\)

And point \(B\) and \(C\) are on the perimeter of circle \(c_1\)

\(\therefore\) segment \(\bar{AC}\) is equal to \(\bar{AB}\)

And point \(A\) and \(C\) are on the perimeter of circle \(c_2\)

\(\therefore\) segment \(\bar{BC}\) is equal to \(\bar{AB}\)

But CA was also proved equal to AB; therefore each of the straight lines CA, CB is equal to AB.

As stated:

\(\therefore\) \(\bar{AC}\) is also equal to \(\bar{BC}\)

\(\therefore\) the three segments \(\bar{AB}\), \(\bar{BC}\), \(\bar{AC}\) are equal to one another.

\(\therefore\) the triangle \(\triangle{ABC}\) is equilateral; and it has been constructed on the segment \(\bar{AB}\).

Todo

demonstrate the second triangle

dependencies