A. C. Twining—Euchid’s Doctrine of Parallels. 337 
For take any line CO making an acute angle A CO with 
. From any point Gin the line drop G F perpendicular to 
CR. Bisect C F in D, and erect DI perpendicular to C R 
and meeting CO inI. In FG, take FH, HK, each equal to 
Fie. 3. 
DI, and join DH, 1H,IK. Because the sides C D, DI, are 
equal to the two D F, F H, and the included angles right angles 
the triangles CID, DH F, are equal and alike in every respect 
(41.), and the angle CID equal to DHF. But O is 
adjacent to DIC at the point I in the straight line C O, and 
similarly D HG is adjacent to D H F,—consequently (13.1.) 
O and D HG are equal angles. 
The angles DCI, CID, of the triangle, right angled at D, 
must together (28. 1.), either be equal to or less than a right 
the two H K, HI of another triangle, each to each, and the con- 
tained angles, are right angles, the triangle HI K is equal and 
alike in every respect to I H D, and also, therefore, to DCI. 
Therefore the angles HI K, DIC together equal a right angle, 
and the three angles CID, DIH, HIK together equal two 
right angles, ‘Therefore (14.1.) CI, IK make one and the same 
Straight line. Also the two angles FC K and F KC of theright 
angled triangle C F K are together equal to aright angle. There- 
fore it has been shown that if the oblique angles of the right 
angled triangle C D I are together equal to one right angle, then CI 
produced will pass througb the point K, making a right angled 
triangle C F K, which, equally with C DI, has its oblique angles 
together equal to a right angle. Produce CF to J, making 
B 
