PRESENTMENT AND PROOF IN (JEOMETRY. 3I9 



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" It looks like the in-centre of it." " Can we prove it?" " Yes 

 (after thoug'ht). because the antiparallels make the angle A with 

 a so that OP bisects RPQ, and so with the others." " Very 

 well ; now can we prove that DEF and PQR are on the same 

 circle?" Here I had to help them out. By antiparallels, 

 DPQE is cyclic, so is QERF ; here, then, are two circles with 

 two points in common : if they have a third, they are identical. 

 But by equality of the angles marked ( FP^FB, being radii), 

 F is concyclic with DPE ; hence we have once more the Ortho- 

 centric Twin-point circle substantially proved (the other three 

 points being easily brought in), and we recognise it as the 

 laringer together of parallels and antiparallels into one bond of 

 symmetry. 



Now we have seen that S and K correspond to O and G ; 

 AO, BO, CO are perpendicular to the parallels, therefore AS, 

 BS, CS are per])endicular to the antiparallels. AG, BG, CG 

 bisect the ])arallels ; therefore AK, I'lv. CK bisect 

 the antiparallels. Then, since pairs of antiparallels to 

 two sides make equal angles with the third side, their perpendi- 

 culars will do so also; hence SA and SB, being perpendicular 

 to lines making the angle C with c, form a symmetrical triangle 

 with the angle 90° — C. at the base; SB and SC behave similarly ; 

 hence SA, SB, SC are equal, and we have the Circum-circle 

 from a new point of view. 



Again, owing to this property, through every point within 

 the triangle there pass three antiparallels, making symmetrical 

 triangles on the sides (I refer again to the tracing-paper drill). 

 But as K is a point, and the only point, where all these three 

 antiparallels are bisected, we see that K is a point, and the only 

 point, from which as centre a circle can be drawn cutting the 

 sides of the triangle at the ends of three diameters. Now, as 

 these diameters are antiparallels to the sides, it follows (from 

 the cyclic property) that the other chords across the triangle 

 are parallel to the sides. This, then, is a simple proof of what 

 is called the Cosine circle — I should prefer to call it the Anti- 



