338 A. @. Twining—Euchid's Doctrine of Parallels. 
“gegen! as RZ, greater than C A, through the extremity 
with CR, RZ, aright angled triangle whose oblique angles at C 
the line CI produced has met A B and crossed it, and must so 
meet and cross in case the angles of C DJ are together two right 
angles as first supposed. But, if C I produced does not pass, as 
above, through K then the angles of C DI cannot be equal to 
two right angles, and therefore they must be less, since 
(28 a, 1) they cannot be greater. 
Suppose then that the angles of CDI are together less than 
two right angles. Then, the construction and proof remaining 
as before, the angles of the triangle DF H are less than two 
right angles, and since the angles of DI H cannot be greater 
than two right angles, the four angles of the quadrilateral I F 
are less than four right angles, and each of the equal angles at 
I and H is less than a right angle. Therefore [H K is greater 
than the angle IHF or its equal HID. Draw H P equal to 
ID, or HK, at the angle I H P equal to HID, which is less than 
a right angle—and therefore I H P falls within the angle IH 4. 
Join IP; then (4. 1.) the angle HPI equals HDI, and 1s 
accordingly acute. It may be supposed either that P shall fall 
within the triangle I H K, as at p, or else in the side I K, as at 
p’, or otherwise outside, as at P. Supposing it at p, let Hp be 
produced to meet IK in p’. Then Hp’ is greater than Hp oF 
its equal H K, and consequently the angle Hp’ K is less than 
H Kp’ (19.1) being subtended by the less side. But HKI's 
acute (17.1) because [HK is obtuse. Much more, then, 18 
H p’ K acute and its adjacent angle H p'I obtuse; and yet more 
is the exterior angle (16.1) Hp I obtuse—which is contrary '0 
the construction, as already proved. Therefore P does not fall 
within THK; and, similarly, it cannot fall in IK, as atP- 
Therefore P must fall outside, and make the angle H I P greater 
than HIK. But because, by what has been shown, DI G less 
