This solution is equivalent to a construction using only a ruler and unmarked
straight edge only. Compasses having been used once to draw the original circle.
Let O be the centre of the circle.
Let A and B be the ends of the given diameter.
The solution is in two parts. In the first, the harmonic construction is used to draw a line parallel to AB which cuts the circle twice.
In the second, these intersections are joined to A and B.
These lines intersect at a point on a line perpendicular to AB through O.
Let P be an arbitrary point on the circle. Extend AP to Q.
Join BP, BQ and OQ.
Let BP and OQ meet at R.
Let AR and BQ meet at S.
By the harmonic construction, PS is parallel to AB.
Let it meet the circle at T and P.
By symmetry, the intersection of AT and BP is on a line through O perpendicular to AB.
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