Geometric Construction with the Compass Alone

Let two points A and B belong to a circle with center O. Bisect the two arcs of the circle defined by the points A and B.

Solution

Start with completing two parallelograms (Problem #6): ABOC (C is opposite to B) and OABD (D is opposite to A). Draw two circles with radius r = AD = BC: one centered at C, another at D. Let E be one of the intersection points. Finally, circles with the radius OE and centers at C and D intersect at the midpoints F and G of the two arcs AB.

Note that the arcs AB do not have to be actually drawn.

Proof

Let R be the radius of the original circle. By construction, the points C, D and O are collinear (lie on the same straight line.) The same is true for the points F, G and O. Furthermore, OF is perpendicular to CD. All we have to show is that OF = R.

For a parallelogram XYZW we have the following identity

which I'll apply to the parallelogram ABOC to get r² = 2AB² + R². Since the triangle COE is right-angled, r² = CE² = EO² + CO², by the Pythagorean Theorem. Therefore, OE² = AB² + R². However, the triangle COF is also right-angled wherefrom OC² + OF² = CF² = OE². Combining the last two identities we get OF = R. Q.E.D.