proof for construction of 60 degree angle I have a construction from the game Euclidea, puzzle 4.2:

The puzzle is given point $A$ and line $\overleftrightarrow{BC}$ (just the line -- neither point is given), construct a 60 degree angle with the line through the point (shown in orange). I came up with this construction to satisfy a goal in the game for performing the construction with a minimum number of elementary steps.
The construction is:


*

*Draw $\odot A$, of arbitrary radius large enough to intersect $\overleftrightarrow{BC}$.

*Draw $\odot C$.

*Draw $\odot D$.

*Adding $\overleftrightarrow{EA}$ forms a 60 degree $\angle ABC$, as if by magic.


What's the proof?
 A: As a convention, let's label the circles with two points - in order, the center and one point on the circle. The steps of the construction, with this in mind:
1) Arbitrary large enough circle centered at $A$. Let $C$ be the intersection of this circle with the given line $\ell$.
2) Circle $CA$. Let $D$ be the intersection of circles $AC$ and $CA$, on the opposite side of $\ell$ from $A$. Let $F$ be the intersection of circle $CA$ and $\ell$.
3) Circle $DF$. Let $E$ be the intersection of circle $DF$ and circle $AC$, outside circle $CA$.
4) Line $EA$ forms the desired $60^\circ$ angle with $\ell$.
Yes, I named a point that you didn't bother naming. We used it, so we should name it.
So, why does this work? Note that $\triangle ACD$ is equilateral, so $\angle CDA=60^\circ$. As such, a $60^\circ$ rotation $\rho$ around $D$ takes $C$ to $A$. Because of that, it also takes circle $CD$ to circle $AD$. Taking the intersections of these circles with circle $DF$ (fixed under the rotation, because its center is fixed), $\rho(F)=E$. Then $\rho(\ell)=\rho(\overleftrightarrow{CF})=\overleftrightarrow{\rho(C)\rho(F)}=\overleftrightarrow{AE}$. The angle between a line and its image under the $60^\circ$ rotation $\rho$ must be $60^\circ$. Done.
