SIMSON. 157 



the lines drawn cutting its circumference and meet- 

 ing a given straight line shall have their segments 

 within and without the circle in the same ratio. This, 

 though a beautiful proposition, is one very easily 

 demonstrated, and is, indeed, a corollary to some of 

 those in the ' Elements.' But Dr. Simson prefixes a 

 lemma: that the line drawn to the right angle of a 

 triangle from the middle point of the hypotenuse, is 

 equal to half that hypotenuse. Now this follows, if 

 the segment containing the right angle be a semi- 

 circle, and it might be thought that this should be 

 assumed only as a manifest corollary from the pro- 

 position, or as the plain converse of the proposition, 

 that the angle in a semicircle is a right angle, but 

 rather as identical with that proposition ; for if we 

 say the semicircle is a right-angled segment, we also 

 say that the right-angled segment is a semicircle. 

 But then it might be supposed that two semicircles 

 could stand on one base : or, which is the same thing, 

 that two perpendiculars could be drawn from one 

 point to the same line; and as these propositions had 

 not been in the elements, (though the one follows 

 from the definition of the circle, and the other from 

 the theorem that the three angles of a triangle are 

 equal to two right angles), and as it might be supposed 

 that two or more circles, like two or more ellipses, 

 might be drawn on the same axis, therefore the lemma 

 is demonstrated by a construction into which the 

 centre does not enter. Again, in applying this lemma 

 to the porism (the proportion of the segments given 

 by similar triangles), a right angle is drawn at the 

 point of the circumference, to which a line is drawn 

 from the extremity of a perpendicular to the given 

 line ; and this, though it proves that perpendicular to 

 pass through the centre, unless two semicircles could 

 stand on the same diameter, is not held sufficient; but 

 the analysis is continued by help of the lemma to show 

 that the perpendicular to the given line passes through 



