# 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:

1. Draw $$\odot A$$, of arbitrary radius large enough to intersect $$\overleftrightarrow{BC}$$.
2. Draw $$\odot C$$.
3. Draw $$\odot D$$.
4. Adding $$\overleftrightarrow{EA}$$ forms a 60 degree $$\angle ABC$$, as if by magic.

What's the proof?

• How is circle D constructed? – fleablood Jan 31 at 4:32
• @fleablood Circle D is centered on the intersection of circles A and C, through the intersection of circle A with line BC. I've updated the image to label that later intersection F. – Phil Frost Jan 31 at 15:18

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.

• I updated the image to label F for clarity – Phil Frost Jan 31 at 15:14