Commentary on Proposition III.25¶
Back to Proposition III.25.
It will be remembered that Simson takes first the case in which the angles ABD, BAD are equal to one another, and then takes the other two cases together, telling us to “produce BD, if necessary.” This is a little shorter than Euclid’s procedure, though Euclid does not repeat the proof of the first case in giving the third, but only refers to it as equally applicable.
Campanus, Peletarius and others give the solution of this problem in which we take two chords not parallel and bisect each at right angles by straight lines, which must meet in the centre, since each contains the centre and they only intersect in one point. Clavius, Billingsley, Barrow and others give the rather simpler solution in which the two chords have one extremity common (cf. Euclid’s proofs of III. 9, 10). This method De Morgan favours, and (as noted on III. 1 above) would make III. 1, this proposition, and IV. 5 all corollaries of the theorem that “the line which bisects a chord perpendicularly must contain the centre.” Mr H. M. Taylor practically adopts this order and method, though he finds the centre of a circle by means of any two non-parallel chords; but he finds the centre of the circle of which a given arc is a part (his proposition corresponding to III. 25) by bisecting at right angles first the base and then the chord joining one extremity of the base to the point in which the line bisecting the base at right angles meets the circumference of the segment. Under De Morgan’s alternative the relation between Euclid III. 1 and the Porism to it would be reversed, and Euclid’s notion of a Porism or corollary would have to be considerably extended.
If the problem is solved after the manner of IV. 5, it is still desirable to state, as Euclid does, after proving AE, EB, EC to be all equal, that “the circle drawn with centre E and distance one of the straight lines AE, EB, EC will also pass through the remaining points of the segment” [III. 9], in order to show that part of the circle described actually coincides with the given segment. This is not so clear if the centre is determined as the intersection of the straight lines bisecting at right angles chords which join pairs of four different points.