Commentary on Proposition III.33¶
Back to Proposition III.33.
Simson remarks truly that the first and third cases, those namely in which the given angle is acute and obtuse respectively, have exactly the same construction and demonstration, so that there is no advantage in repeating them. Accordingly he deals with the cases as one, merely drawing two different figures. It is also true, as Simson says, that the demonstration of the second case in which the given angle is a right angle “is done in a roundabout way,” whereas, as Clavius showed, the problem can be more easily solved by merely bisecting AB and describing a semicircle on it. A glance at Euclid’s figure and proof will however show a more curious fact, namely that he does not, in the proof of the second case, use the angle in the alternate segment, as he does in the other two cases. He might have done so after proving that AD touches the circle; this would only have required his point E to be placed on the side of AB opposite to D. Instead of this, he uses III. 31, and proves that the angle AEB is equal to the angle C, because the former is an angle in a semicircle, and is therefore a right angle as C is.
The difference of procedure is no doubt owing to the fact that he has not, in III. 32, distinguished the case in which the cutting and touching straight lines are at right angles, i.e. in which the two alternate segments are semicircles. To prove this case would also have required III. 31, so that nothing would have been gained by stating it separately in III. 32 and then quoting the result as part of III. 32, instead of referring directly to III. 31.
It is assumed in Euclid’s proof of the first and third cases that AE and FG will meet; but of course there is no difficulty in satisfying ourselves of this.