Proposition 1, Problem

On a given finite straight line $({\color{#000} \rule[-0.5em]{1cm}{2pt} })$ to describe an equilateral triangle.

Describe the blue circle, whose radius is the black line (post. 3.).

Describe the red circle, whose radius is the black line (post. 3.).

Draw lines ${\color{#FFd700} \rule[-0.5em]{1cm}{2pt} }$ and ${\color{#D32F2F} \rule[-0.5em]{1cm}{2pt} }$ (post. 1.).

Then will the triangle $( {\color{#000} \rule[-0.5em]{1cm}{2pt} }, {\color{#FFd700} \rule[-0.5em]{1cm}{2pt} }, {\color{#D32F2F} \rule[-0.5em]{1cm}{2pt} } )$ be equilateral.

For

={\color{#000} \rule[-0.5em]{1cm}{2pt} } = {\color{#FFd700} \rule[-0.5em]{1cm}{2pt} }

(def. 15.); and

={\color{#000} \rule[-0.5em]{1cm}{2pt} } = {\color{#D32F2F} \rule[-0.5em]{1cm}{2pt} }

(def. 15.),

∴=\therefore {\color{#D32F2F} \rule[-0.5em]{1cm}{2pt} } = {\color{#FFd700} \rule[-0.5em]{1cm}{2pt} }

(axiom. 1.);

and therefore we constructed the required equilateral triangle.

Q.E.D.

I made a Geogebra of this proposition: