Compass equivalence theorem
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Through Geometry, Straightedge and compass construction and Circle
Construction
The following construction and proof of correctness are given by Euclid in his Elements. Although there appear to be several cases in Euclid's treatment, depending upon choices made when interpreting ambiguous instructions, they all lead to the same conclusion, and so, specific choices are given below.
Given points A, B, and C, construct a circle centered at A with radius the length of BC (that is, equivalent to the solid green circle, but centered at A).
- Draw a circle centered at A and passing through B and vice versa (the red circles). They will intersect at point D and form the equilateral triangle △ABD.
- Extend DB past B and find the intersection of DB and the circle ◯BC, labeled E.
- Create a circle centered at D and passing through E (the blue circle).
- Extend DA past A and find the intersection of DA and the circle ◯DE, labeled F.
- Construct a circle centered at A and passing through F (the dotted green circle)
- Because △ADB is an equilateral triangle, DA = DB.
- Because E and F are on a circle around D, DE = DF.
- Therefore, AF = BE.
- Because E is on the circle ◯BC, BE = BC.
- Therefore, AF = BC.
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Alternative construction without straightedge
It is possible to prove compass equivalence without the use of the straightedge. This justifies the use of "fixed compass" moves (constructing a circle of a given radius at a different location) in proofs of the Mohr–Mascheroni theorem, which states that any construction possible with straightedge and compass can be accomplished with compass alone.
Given points A, B, and C, construct a circle centered at A with the radius BC, using only a collapsing compass and no straightedge.
- Draw a circle centered at A and passing through B and vice versa (the blue circles). They will intersect at points D and D'.
- Draw circles through C with centers at D and D' (the red circles). Label their other intersection E.
- Draw a circle (the green circle) with center A passing through E. This is the required circle.
There are several proofs of the correctness of this construction and it is often left as an exercise for the reader. Here is a modern one using transformations.
- The line DD' is the perpendicular bisector of AB. Thus A is the reflection of B through line DD'.
- By construction, E is the reflection of C through line DD'.
- Since reflection is an isometry, it follows that AE = BC as desired.

